HC Verma Physics Part 1 Solutions for Chapter 9 Centre of Mass, Linear Momentum, Collision

5/21/2012

Download HC Verma Solutions for HC Verma Physics Part 1 Solutions Chapter 9 Centre of Mass, Linear Momentum, Collision solved by our expert teachers. We have curated solutions for all questions of chapter 9. You can download the Solution of HC Verma - Centre of Mass, Linear Momentum, Collision and prepare for your upcoming competitive examinations.

Solution of HC Verma Concept of Physics -1- Chapter 9 Centre of Mass, Linear Momentum, Collision

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306 Comments

Chandan badtya link

9/14/2012 07:09:04 pm

This is a good process to improve my knowledge about problems of physics

Lady Gaga link

10/28/2014 09:37:06 pm

yes

sassad

7/25/2015 02:25:43 pm

bhai tu chup bait

advait patole

10/18/2012 03:26:50 am

not understood solution of center of mass 22question {exercises}

Admin

10/18/2012 11:42:15 am

we know that explosion of block is an internal force. there is no external force acting on the body. so here we are using the concept of conservation of momentum which means that momentum before explosion and after explosion remains same. as given in the question that the body was at rest, so the momentum was zero.
here also after explosion we are balancing the momentum so that net momentum becomes zero.
Here two bodies are moving in x and y direction with 10 m/s each. so we calculate its net velocity with vector addition formula. so our answer comes out to be 14.14 m/s with angle of 45 w.r.t x- axis. so in order to balance this momentum third mass should move in opposite direction with the same velocity because mass is equal.
so third mass is 180 angle out of phase with the above resultant and with x- axis it is making the angle of 135.

Pragyan

11/13/2014 01:07:46 pm

can you explain me how to calculate time of flight in question no. 55 of chapter 9 (exercises) of hc verma

yucv

3/16/2019 06:17:46 am

same here

Aman

10/29/2012 10:51:41 pm

please explain Q-31, ch-9

anisha

9/22/2015 12:18:59 am

Conserve the momentum and equate it of both the balls😊

Admin

11/3/2012 03:08:24 pm

This is attempt to provide free education to children who are preparing for engineering entrance examination.
If you are having any doubt regarding solving questions of HC Verma or understanding any topic related to this, you may leave your comment.
we will give your answer within 24 hours.
Thank You

Avinash

11/4/2012 12:20:52 pm

Sir, this site really proved to be very helpful for me..!!! Very nice solutions and well explained too..!!! Please do something for the short answer type questions of HCV..!! thank you.

Admin

11/4/2012 07:55:50 pm

Dear Avinash you can also ask "short answer type question" in which you are facing problem.

Karan Muvvala

11/7/2012 07:47:28 pm

sir can u explain me q 24 and how to find average force delivered

Admin

11/7/2012 09:28:01 pm

Dear Karan,
In this question do not get confused with the average force. It is just asking the total force acting on the roof.
we know that force = rate of change of momentum
we know that rate of change of momentum = rate of initial momentum - rate of final momentum
Here final momentum = 0 because hailstorm do not rebound.
so here we will only calculate the rate of initial momentum
for this we have calculated the mass of one hailstorm using m = density*volume.
it is given that hailstorm is falling at the rate of 2000 hailstone per second per unit area.
so calculate the mass of 2000 hailstone in 100 area by multiplying mass of one hailstone with 2000 and 100.
now multiply the obtained answer with velocity i.e 20 and we get the answer.

Rajat

11/15/2012 08:42:43 pm

pls explain...
Ques number 42 of chapter 9.

Rachana

7/13/2017 07:12:33 am

Bullet stokes the block with a speed 500m/s.and the bullet comes out of the block with the speed 100m/s.Let us take that this happens in a very short time and conserve momentum,
mu=mv+MV,
here m is the mass of the bullet ,M is the mass of block.u,v are initial and final velocities of the bullet,V is the velocity of the block after bullet comes out of the block.
(20/1000)500=(20/1000)100+10V
by solving this we get,
V=0.8
V^2 -U^2 =2as
By this we get a=1.6
Coefficient of friction=a/g=0.16

Admin

11/16/2012 11:50:05 am

Your answer will be uploaded on 18th Nov.
Please wait for few hours because there is some technical problem in the Website. we are trying to serve you as fast as possible.

Rajat

11/16/2012 02:13:51 pm

Its absolutely ok! I'll wait for the answer...

Admin

11/17/2012 06:05:48 pm

Dear Rajat,
In this question a Bullet is colliding with block which is kept at rest.
Now whenever you are solving these type of questions, follow the simple rule i.e. take both the objects as one system. So, when we consider these bullet and block as one system then there is no external force acting on the system. So when there is no external force then we can say that there is conservation of momentum.
So after applying the equation of momentum conservation we have calculated the final velocity of block.
we know the fact that change in kinetic energy = work done on the block.
work done = force * displacement
through this we get another equation.

arjun

5/27/2013 11:13:21 pm

sir pls explain why enegry conservation eq give wrong ans in this question i.e
k.e bullet+ke of blocki.e o =ke of bullet final+ke of block final +wok done by friction

if acc to sol. why in change in k.e the intial kinetic energry is taken as 0 (as bullet has intial velocity)

Mainak Pal

11/26/2015 10:35:10 am

but what about the frictional force???

Rajat

11/17/2012 08:36:58 pm

Got it ! thank you..
I have one more thing to ask...maybe its a stupid question but i want to knw what is 'angular frequency' in SHM. There is no circular motion involved here, so what it means?

Admin

11/18/2012 12:00:58 am

Dear Rajat,
Every SHM motion can be represented in the form of circular motion in which radius is taken as the amplitude of the SHM motion.
Now imagine that a circle of radius A is made with the center at (0,0) with X and Y as Horizontal and Vertical axis respectively.

Now consider the example of Sine wave of Amplitude (A). when a particle starts moving on a sine wave, its vertical distance changes continuously ranging from +A to -A.
Now I am explaining its circular motion. Consider the circle of radius A. Please see the figure by going to below link.
http://tinyurl.com/b5suoet
Here we can see that the radius AB is moving in the anti-clockwise
direction starting from X axis. After time (t) the radius AB is making an angle @ with the horizontal where @= w*t where w = angular velocity or angular frequency. Now resolve the Radius AB along the Y- axis and you will get the vertical distance of particle from the mean position = A*[email protected]
where angle @ = w*t and w = angular frequency which you were asking about.

Rajat

11/18/2012 12:37:28 am

Sir you explain very well...now i understood the real meaning of angular freq ! Thank you for the help....

m k rao

11/18/2012 07:43:37 pm

sir can u explain q36 step 11 and q37

Admin

11/18/2012 08:44:25 pm

Dear M K Rao,
In question number 36,
It is given that one block is colliding with another block.
In part A - Maximum loss in Kinetic Energy will occur when there will be pure inelastic collision.
first we have calculated the common velocity after inelastic collision. so total loss of energy = Initial Kinetic Energy of both
blocks - Final Kinetic Energy of both blocks.
For Part B -
In this question it is given that the loss is half of the maximum Kinetic energy loss.
let us assume that v1 = velocity of 1st block after collision
v2 = velocity of 2nd block after collision
now through the above given statement we form one equation.
Now from the law of conservation of momentum we form another equation i.e
4 = 2*v1 + 2*v2
Now we have two variables and two equations and through this we can calculate v1 and v2. Through v1 and v2 we may calculate cofficient of restitution.

Admin

11/19/2012 07:33:57 pm

Dear M K Rao,
In Question 37 , it is given that a mass with velocity u is colliding with second mass at rest. After collision the energy becomes 0.2J.
Now we have assumed v1 and v2 as the final velocities of 1st and 2nd block respectively.
Through Law of momentum conservation we have calculated one equation.
Now in order to calculate another equation we are using cofficient of restitution.
Do not get confused with the notations in above solutions. Here e = cofficient of restitution and by mistake we have written a instead of u. Here a= u.
After this we have calculated v1 and v2 in terms of u and e. Then we calculated the relation between u and e.
For maximum value of u, e = 0
For minimum value of u, e = 1

KUNAL

5/9/2015 09:34:09 pm

BUT IN A I M GETTIN 0 AS TOTAL LOSS
AS COFF. OF REST. OS 1

hahaha

9/10/2016 03:59:20 am

(v1 – v2) + ℓ (a1 – u2) = 0  ℓa = v2 – v1 ..(2)
what do mean by this equation sir..
what is l?
i guess you have written the equation of coefficient of restitution..that should be ---- v1-v2+eu=0 but then sir for perfectly inelastic collision the value of e should be 0.

Karan Muvvala

11/20/2012 01:40:00 am

in question 55 i did not understand the last step

Admin

11/20/2012 08:11:29 pm

Dear Karan,
In the solution first we are trying to calculate the angle @ at which the block wil loose contact with the track.
Let D is the point at which the block will loose contact.
We know that centripetal force during circular motion is provided by the component of weight and it is [email protected]
There is some misprints in the diagram. Here ([email protected]) is angle of projectile velocity with the horizontal and @ is the angle between DB and horizontal.
Now using energy conservation law we have calculated the angle @.
There is some misprints in the formula also.
Here S = V cos ([email protected]) * t where V cos ([email protected]) = horizontal component of the velocity. using this you will get S = 1.44 m which will fall near junction.

Raman

11/24/2012 05:19:51 pm

sir please explain me the 59th question ,i wanted to know the equation v1-v2=(u1-v2)is the eqn for coefficient of restituion

Admin

11/25/2012 09:11:13 am

Dear Raman,
Yes v1-v2 = (u1-u2) is the equation for cofficient of restitution.
We know that for perfectly elastic collision coefficient of restitution is unity i.e. 1.
so velocity of seperation = velocity of approach

Admin

11/25/2012 09:16:54 am

Dear Raman,
If you are facing any problem in understanding any concept then you may write to us. we will solve your problem as fast as possible.

Raman

12/1/2012 11:46:25 pm

Please explain me the 3rd and 16th question of chapter 10 ie rotation of HC Verma.Just getting a bit confused

Admin

12/2/2012 10:58:45 am

Dear Raman,
#Solution for Question 3
It is just the graphical method for solving the problem.
You know that θ = ωt
So we made the graph between angular velocity (ω) and time(t)
Always learn the rule that area under curve gives us the product of factors between which the graph is made. For example
1. Area under ω and t gives us the product of ω and t which is θ.
2. Area under velocity and time curve gives the product of velocity and time i.e Distance.
We just made graph according to instructions given in the question and find its area.

#Solution for question 16
We know that surface density is mass/area. So, total mass = surface density x area
Now we take a small area(dA) = 2πrdr
Now we get small mass(dm) = (A+Br) 2πrdr
Moment of inertia(I) = ʃdmr²
I = ʃ(A+Br) 2πrdr x r²
I = ʃ(A+Br) 2πrᶟdr
I = ʃ(2Aπrᶟdr+2Bπr⁴dr)
In integration there is rule that
ʃ Cxᶢ dx = C xᵛ/v = M₂ - M₁
Where v = g+1 and 2 and 1 are the upper and lower limits of x
After integration we get
I = 2Aπ[r⁴/4]+ 2Bπ[r⁵/5]
Now we apply upper limit (maximum value) of r = a and lower limits (minimum value) of r = 0 as shown in the solutions.
After this limits are applied we get the answer

Raman

12/1/2012 11:48:22 pm

And sir I'am just not arriving at the right answer so please do solve the complete 16 question or atleast do write the basic integrating equation

Raman

12/2/2012 12:18:08 pm

Sir the 21st question is really getting me confused.How could the normal force of the block or any of the components of it or even weight can produce torque about the centre of the block.I 've downloaded the sol of HC verma of all chapters and the solution given for this is totally absurd.Please reply at the earliest ....

Raman

12/2/2012 12:22:16 pm

Sir for the 3rd question of rotation ,i know that the area under the curve gives the value of the prod of quantities taken but here we need ang velocity and ang acceleration is given which is not constant and then graph for velocit would be curved and in the sol it is given a straight line.Getting totally confused seeing the solutions

Raman

12/2/2012 12:40:06 pm

Sir sorry about so many doubts but you are really helping me a lot,I'am studying in 10th and just wanted to complete the important chapters of HCV ,so u are my only teacher at the moment.
My doubt was in question 23 of rotation which was that in solutions I is calculated by integration .Can't we do it by calculating I about the line which can be considered as the diametric line about the centre and then by parallel axis theorem find I about the line through the edge.Sir please provide me the solution by this method and if you could please supply me the moment of inertia results for various objects including box,plates about various axis of rotations.

Raman

12/2/2012 12:49:09 pm

Sorry sir but i've got the answer of ques 23 by that method ,i just took the wrong Moment of inertia .......Please do reply for the other questions

Admin

12/3/2012 10:03:56 am

# Solution of question number 21
Here the language of the question is tricky. Please read the statement again that find the torque due to normal force acting on the block.
Please see in the diagram that there are three forces that are normal to the block. These are
1. mgcosθ
2. R = Reaction force
3. mgsinθ

Torque due to first two forces i.e mgcosθ and R is zero because they are passing through the centre
but mgsinθ is not passing through the centre and there will be torque due to this force which is give
by torque = mgsinθ(a/2)

# Solution for question number 3 –
Dear do not get confused with the angular acceleration.
We know that ω₂ = ω₁+ αt
Where ω₂ = final angular velocity
ω₁ = initial angular velocity
α = angular acceleration
t = time
Now consider the first 10 seconds
ω₁ = 0
α = 4 rad/s²
t = 10 seconds
ω₂ = 0 + 40 = 40 rad/s
Now I am clearing your doubt regarding curve. The given equation
ω = ω₁+ αt
is a straight line equation because this is a graph between ω and t and the power of time (t) is one in the equation. That’s why ω is a straight line. You will study it in 11th class in straight line chapter.
Now draw the straight line joining 0 and 40 in the graph.
Now considering next 10 seconds –
Here the body is rotating constantly at ω = 40 rad/s.
Now draw a straight line parallel to X- axis for these 10 seconds because ω is not changing.
Now considering another 10 seconds -
It is given that it is brought to rest
Here ω₂ = 0
ω₁ = 40
t = 10
After solving we get α = -4 rad/s².
It will again be the straight line.
Now join 40 and 0 in the graph.
Here your diagram is ready.

Raman

12/3/2012 11:55:28 am

sir in your explaination for question 21 the thing I'am not understanding is that how can mgsin0(0 is for theta) produce a torque because it is like mgcos0 is a component of weight so it will also pass through the centre ,is there some reason because of which sine component of weight shifts and is not acting on the centre.If there is some reason sir please explain me...waiting for your answer

Admin

12/4/2012 09:58:46 am

Dear Raman,
Let us read the question carefully. It is given that the block is moving on a rough surface which means that there will be friction acting on the body.
Secondly it is given in the question that the block is moving with constant speed
i.e v = constant.
This means that acceleration is zero i.e. a = 0
It is only possible if all the forces are balanced or net force in the moving direction is zero.
Now the normal force R = mgcosθ
mgsinθ = friction force(f)
which means that friction force is balancing the component mgsinθ.
And the torque is generated due to the friction force(mgsinθ) and not due to weight.

Admin

12/4/2012 10:01:02 am

Dear Raman,
Please post your comments in their respective chapters.

Raman

12/3/2012 07:06:09 pm

Sir please explain the 44 question of rotation.The diagram in the solutions given is totally confusing.If possible please explain the what forces and how are they mentioned in context of the diagram..

Raman

12/3/2012 07:07:43 pm

Sir please explain the 44 question of rotation.The diagram in the solutions given is totally confusing.If possible please explain what forces will act and how are they mentioned in context of the diagram..

Admin

12/4/2012 10:19:31 am

Dear Raman,
Here two forces are acting in two different planes.
first force which will balance the weight i.e F1 and F2 is acting in the plane of the door.
The other force is the reaction of the force which is supporting the door is acting in the plane perpendicular to door..
The second force can act at any angle because door is hinged and is able to rotate any where but it will act only in the plane perpendicular to the door.
Let it be s. Now we resolve this force in the horizontal and verticle direction in that plane.
we get N and R along X and Y direction respectively.
Now N1 and N2 are the forces along X direction on hinge 1 and 2 respectively.
Similarly, R1 and R2 are the forces along Y direction on hinge 1 and 2 respectively.
As we can see that there is no net or external force acting in the plane perpendicular to door, so these forces will balance each other.

Raman

12/3/2012 08:59:49 pm

Sir please explain the in depth concept of questions 56,57,58 of rotation that is how does torque remains 0 and what is the basic logic behind these type of questions.
Though i have got the answers write for these all but still explain me if there are some other important things related with these and basically about the practical aspect of this question that is how does and in what direction does the omega of platform change to conserve momentum and other such things

Admin

12/4/2012 10:39:42 am

Dear Raman,
These are the questions of conservation of angular of angular momentum. Generally in these types of questions a situation is given.
For example - In question number 56 it is given that given that a boy is catching a ball moving at velocity (v). Now whenever you are solving these type of questions, follow the simple rule i.e. take both the objects as one system. Now check that both the bodies should not be carrying any force. we can see that before the ball was catched, the ball was moving with constant velocity v with zero acceleration and also the the boy was standing at rest then we can see that there is no external force acting on the system. So when there is no external force then we can say that there is conservation of momentum.
Also the conservation of momentum would have been violated if the ball would have been moving with some acceleration because it is carrying some force now.
so before solving these questions make sure that all the bodies of your system should not be accelerating otherwise you will not be able to apply conservation of momentum.

Raman

12/4/2012 01:01:10 am

sir explain me the questions 59 to 63 of rotation.I'am getting confused about when to take linear momentum and ang momentum and sometimes simultaneously linear and ang mom and other such things.Please explain the basic concept behind all this so that my basics get really clear about ang momentum and i can deal with such problems

Admin

12/5/2012 10:25:27 am

Dear Raman,
For knowing the application of angular and linear quantities.
Always follow the simple rule - A Force acting on the body has the two functions to perform –
1. It will move or accelerate the body in the direction of the force irrespective of the point of application of the force.
2. It will rotate the body about the centre of mass if the force is acting at a point other than the centre of mass.
So it means that everything is the combination of linear and angular motion which means that the body is moving as well as rotating where the linear quantities like velocity, acceleration etc are found by linear momentum or force and angular quantities like ω, α are found by angular momentum or torque.
Now I am going to explain you the question on this simple rule.
Solution for question number 59
Part a – To calculate the velocity –
Here do not think that if the force is acting on the end then it will only rotate. It will rotate and also it will move the body in the direction of force.
Whenever the question is asking about the centre of mass then always use linear quantities because the centre of mass will not rotate about itself.
So here we are using linear momentum equation to calculate the velocity.
Part b -
calculate angular speed. We simply use angular quantity.
Part c- Calculating kinetic energy –
Here there are two kinetic energies. First is due to linear motion and second is due to rotational motion.
Similarly try to solve other questions with the rule explained above. If you still feel uncomfortable then tell us and we will give you the detailed explanation about the other solutions.

Raman

12/4/2012 11:33:27 am

thanks a lot sir about the 21st question,i just didn't read the ques carefully which mentioned that velocity is uniform ie a=0 and therefore friction balances mgsin0.

rohan kapadnis

12/5/2012 12:13:08 am

sir can u plz explain me the 11th and 12th question of centre of mass

Admin

12/5/2012 08:32:58 pm

Dear Rohan,
Solution for question 11
For question 11 we have used the simple formula
Y₀ (A₂ + A₁) = A₂Y₂ + A₁ Y₁
Where A₁ = area of inner semicircle is taken as negative because it is being removed.
A₂ = area of outer circle
Y₂ = distance of centre of mass of A₂ from the diameter
Y₁ = distance of centre of mass of A1 from the diameter
Whenever you are solving the questions in which you have to find the centre of mass of composite geometries, then first divide the geometry in the standard areas whose centre of mass of known to us. For example if you want to find the centre of mass of body having area A. Then first we divide the body in the standard areas like rectangle, circle etc whose centre of mass is known to us. Let us divide A into A₁, A₂, A₃ etc having centre of mass Y₁, Y₂, Y₃ etc.
Use the formula
Y A = (Y₁A₁+Y₂A₂ +A₃Y₃)
Y = Centre of mass of body A
Y₁ = Centre of mass of area A₁
Y₂ = Centre of mass of area A₂
Y₃ = Centre of mass of area A₃
If area is removed from the geometry then the value of removed area is taken as negative. For example in the question there is a semicircle of radius R₂ in which the semicircle of radius R₁ is removed. Therefore, we use area of semicircle of radius R₁ as negative.

Solution for question number 12 –
In this question, it is given that two persons are standing at some distance on the boat. So it is having centre of mass at some point which is calculated as 1.87 m as given in the solutions.
Now if the two persons will come towards each other then the centre of mass will shift towards the centre which is 2 m. Now the shift in the centre of mass = 2-1.87 m = 13 cm towards right. But we know that there is no external force acting on the body then centre of mass will move in the opposite direction in order to cancel its effect.

Raman

12/6/2012 12:08:47 am

Sir please i have doubts in the 85th and 86th questions.Sir in the 85th question ho wto take the directions of the linear and angular momentum ,i mean that when will they be opposite to each other because in ques 86th thefinal linear and ang momentum being opp to each other are taken as opposite.I just didnt understand hpw will one decide the direction .Sir please explain me the 85th question in detail,i got to the answer but now some things are confusing me and please write the full equation.

VIVEK

12/25/2012 03:48:37 pm

sir please explain ques-29 chap-9(exercise)

admin

12/25/2012 08:40:08 pm

Dear Vivek,

In the question it is given there are two blocks. one is small and one is large.
small block is moving on large block at constant velocity. as both the blocks are non accelerating, then we can say that no external force is acting on them.
if external force acting is zero, then we can say that momentum of the system will remain conserved.
so, applying law of conservation of momentum w get,
mv + M x 0 = (M+m)v1
v1 = mv/(m+M)

Ankit link

9/30/2014 03:36:28 pm

how can we say that there final velocities will be equal??

devesh yadav

12/25/2012 04:55:22 pm

Sir please explain me the question no 44 from chapter 9...

admin

12/25/2012 08:17:52 pm

Dear Devesh,
in this question a ball falls from height h on an inclined plane and the collision is perfectly elastic.
Now, as the ball is falling from height h, then velocity of collision of ball = √(2gh)
also as the collision is elastic, then velocity after collision = √(2gh)
let ball moves at an angle α with the horizontal after collision.

here we have just used the trajectory equation of projectile which is
y = x tan α – (gx² sec² α)⁄2u²

Now we will just find the respective variables in the above equation.
now please see the above example in the solutions. we are going to solve the question with respect to the diagram given in the solutions.
y = height from the point of base where the ball is landing = l sin θ
x = distance between point of projectile and point of landing of ball = l cos θ
α = angle between projectile velocity and horizontal = (90 - 2θ)
now we put these values in the above equation and solved the value of l.
if you are facing problem in solving any variable then you may write to us.

Kushal

12/21/2013 01:36:09 am

why u are taking y = l sinθ ... because in soln there is y= - lsinθ

rahul

4/2/2015 06:22:08 pm

Why alpha= 90-thita

Rajat

1/23/2013 03:01:07 am

A man starts walking on wedge which is placed on smooth surface. When the man reaches the top most position we have to find the displacement of wedge wrt earth.
they have conserved the momentum in horizontal direction. Please explain why not in vertical direction??

Admin

1/23/2013 06:34:43 pm

Dear Rajat,
This is so because we want to calculate the distance in the horizontal direction. Also the wedge will not move in the vertical direction.
Also in moving in the vertical direction we need to apply the force against the gravitation force which would violate momentum conservation. you may see the example for this on page number 152, question 6.

shalu

1/24/2013 10:52:52 pm

sir please explain me qestion no 54 and 59 and center of mass of continous bodies

Admin

1/25/2013 10:56:29 pm

Dear Shalu,
Explanation for question 54-
In this question we asked to find the velocity of A with which A should move so that after colliding with B, the Block B reaches the head of man.
In this question it is given that the platform is smooth and there is no external force. so, there will be no friction which means that we can apply conservation of momentum and conservation of energy. Just with the help of these two theorems we are able to solve the question.
Let initial velocity of A which is to be calculated = u1
velocity of A just before collision = v1
velocity of A just after collision = v
velocity of B before collision = 0
velocity of B after collision = √(2gh) as explained in the solution
velocity of B at man's head = 0
step 1 -
we first apply conservation of energy from starting point to the point before collision and we calculate v1 in terms if u1.
step 2-
we apply conservation of momentum before and after collision and form one equation in terms of v1 and v.
step 3-
we calculate conservation of energy before and after the collision and form another equation in terms of v1 and v.
step 4-
we calculate the value of v1 and v.
step 5 -
after getting v1, we calculate u1.

Explanation of question 59
Here also block of 2 kg is moving towards block of mass 4 kg.
initial velocity of block 2 kg = 1 m/s
velocity of 2 kg block just before collision = u1
velocity of 2 kg block just after collision = v1
Velocity of 4 kg block just after collision = v2
velocity of 4 kg block before collision = 0
distance traveled by 2 kg block after collision = s1
distance traveled by 4 kg block after collision = s2
step 1-
we first calculate u1 with the help of work energy theorem where change in kinetic energy = work done
here velocity is changing from 1 m/s to u1 m/s and work done by friction = µmgs
where s = distance traveled 16 cm.
step 2 -
we apply momentum equation of 2 kg and 3 kg block just before and after collision.
And form the equation in terms of v1 and v2
step 3 -
we know that the the blocks are colliding elastically then we are using coefficient of restitution and form another equation.
where e = velocity of separation / velocity of approach
and get another equation.
Now we get the value of v1 and v2.
step 4-
we apply work energy theorem on 2 kg block and calculated the value of s1.
we apply work energy theorem on 4 kg block and calculated the value of s2.

center of mass of continous bodies will be uploaded further.

Admin

1/26/2013 11:42:33 am

Dear Shalu,
Here I am going to explain you about the Center of mass of continuous body. It is also same as that of the center of mass of discontinuous body having mass m1,m2,m3.......
center of mass = m1(x1)+m2(x2) +m3(x3)+.......+mn(xn)/(total mass)
Now body becomes continiuous when n becomes infinity or we can say that different particles of mass becomes so small in size and large in number that we cannot cannot add them separately. so, we use integration. That's why the formula becomes
center of mass = (integration( x dm))/(total mass)
You could see the examples given in your books regarding this.

yash

1/25/2013 01:22:16 pm

sir i have doubt in q25.. that how the impulsive forces are taken as zero after long interval of time..

Admin

1/25/2013 11:01:05 pm

Dear Yash,
Yes you are thinking right the force is acting for a small period of time when a ball is in contact with floor and for the rest of time no force is acting.
but in the question they are asking about the average force where we have to take the whole time because
average force = total change in momentum/ total time

pari

2/6/2013 10:13:47 pm

mass=densty *volume
dm=(density)dv
bt in findin cenre of m of semicircul ring why we take mass per unit lenth
in cse of semicircular disc mas per unit area
in hemispher mas per unt volume hw to find the dm of any object plz help me i dnt undrstnd why dm take diff in diff case

Admin

2/7/2013 07:36:19 pm

Dear Pari,
In this derivation our purpose is to find small change in mass i.e. dm
so, whenever you are finding dm you need to see that how the mass is distributed in the body. For example -
1. In case of ring - The mass is distributed along its length. we assume that there is no width and height in ring and mass is only distributed along the length.
2. In case of disc - you can clearly see that the disc is having 2 dimensional body. Although it is having some thickness but its thickness is very small and negligible as compared to length and breadth. so, we can say that the mass of disc is distributed along its area. That's we have taken mass per unit area in this case.
3. In case of hemisphere - It is 3 dimensional body and its mass is distributed along its volume.

ruby

2/6/2013 10:24:30 pm

examp : 2 ,18

Admin

2/7/2013 07:50:46 pm

Dear Ruby,
Explanation for question 2
Here in this question we need to find the center of mass of the rectangular body whose have body is having density as ρ1 and half of the body is having density ρ2.
Now Let us divide the body in two halves. These are
1. This is the plate of length L/2 and having density ρ1.
2. This is the plate of length L/2 and having density ρ2.
As we can see that this is plate so, units of density is kg/m².
so, let us assume that area of both the plates is A.
so, mass of plate 1 = (ρ1)A
mass of plate 2 = (ρ2)A
total mass of the plate = (ρ1)A + (ρ2)A

Now x - coordinate of center of mass of plate 1 from point 0 = L/4
x - coordinate of center of mass of plate 2 from point 0 = 3L/4

x - coordinate of the whole plate =
((ρ1)AL/2 + 3(ρ2)AL/2) / ((ρ1)A + (ρ2)A) = Answer

Admin

2/7/2013 08:04:16 pm

Dear Ruby,
Explanation for question 18
In this question we need to find final speed of two blocks which are connected by the spring.
Here first we will apply conservation of momentum theorem and then conservation of energy theorem to solve the question because no external force is applied on the system.
V = velocity of bigger mass
u = velocity of smaller mass
Let us assume motion towards right to be positive and motion towards left be negative.
we can see that object are moving in opposite direction opposite to each other. M is moving towards right and m is moving towards left. that's why v is negative.
Now apply conservation of momentum theorem,
MV - mv = 0
V = mv/M

Now apply conservation of energy theorem
Initial kinetic energy = energy for spring = (1/2) kx²
Final energy of system when spring acquires natural length =
(1/2)MV² + (1/2)mv²
At natural length x = 0
Now using above two equation we can calculate v and V.

ruby

2/7/2013 12:23:32 pm

plz expln Q:7 i dnt undrstnd

Admin

2/7/2013 08:24:41 pm

Dear Ruby,
Here we have just used the method of vectors to solve it. If your are not familiar with the vectors then we can also use another method to solve it.
V(x) = velocity of center of mass in x direction
u = velocity of particles in the x direction
V(x) = (m1 x u1+ m2 x u2 + m3 x u3 + m4 x u4 + m5 x u5) / (sum of all masses)
V(y) = component of center of mass in the y direction
v = component of velocities of all the respective masses in the y direction
V(y) = (m1 x v1 + m2 x v2 + m3 x v3 + m4 x v4 + m5 x v5) / (sum of all masses)
Net velocity = √( (V(x))² + (V(y))² )

M.y

2/7/2013 12:37:35 pm

Q 11 of excrse

Admin

2/7/2013 08:26:53 pm

Dear M.y,
Solution for question 11
For question 11 we have used the simple formula
Y₀ (A₂ + A₁) = A₂Y₂ + A₁ Y₁
Where A₁ = area of inner semicircle is taken as negative because it is being removed.
A₂ = area of outer circle
Y₂ = distance of centre of mass of A₂ from the diameter
Y₁ = distance of centre of mass of A1 from the diameter
Whenever you are solving the questions in which you have to find the centre of mass of composite geometries, then first divide the geometry in the standard areas whose centre of mass of known to us. For example if you want to find the centre of mass of body having area A. Then first we divide the body in the standard areas like rectangle, circle etc whose centre of mass is known to us. Let us divide A into A₁, A₂, A₃ etc having centre of mass Y₁, Y₂, Y₃ etc.
Use the formula
Y A = (Y₁A₁+Y₂A₂ +A₃Y₃)
Y = Centre of mass of body A
Y₁ = Centre of mass of area A₁
Y₂ = Centre of mass of area A₂
Y₃ = Centre of mass of area A₃
If area is removed from the geometry then the value of removed area is taken as negative. For example in the question there is a semicircle of radius R₂ in which the semicircle of radius R₁ is removed. Therefore, we use area of semicircle of radius R₁ as negative.

Prateek

2/8/2013 12:52:37 am

sir..can u explain ques 27 of chapter 9..i didnt quite get it..then after 1st bullet is fired the car already attains sum velocity...then how to conserve momentum afterwards..when second bullet is fired..
Same funda will apply for ques 28..i think.. so please explain this concept..

Admin

2/8/2013 05:27:02 pm

Dear Prateek,
First I am explaining you where to apply conservation of momentum theorem. Now when ever you are solving these types of questions take all the objects as a system. Now see that these objects should not be carrying any force. This means that either the objects of the system should be at rest or they are moving at constant speed. Theorem of conservation of momentum violate in two cases. These are -
1. If the external force is applied on the system
2. if the objects in the system are moving with acceleration or they are carrying some force.

Now let us come to the question -
Now first we have applied conservation of momentum theorem when first bullet is fired because all the objects of the system were initially at rest.
Now why we applied the conservation of momentum theorem in the second shot because car was moving with constant velocity. This means that car is not carrying any force or no external force is applied. If car would be accelerating then we conservation of momentum theorem would have failed.

Prateek

2/8/2013 01:00:43 am

Sir..in the solution of ques 28 above..they havent considered the velocity which the car attains after first person jumps in momentum conservation..they have solved two cases differently..and then took out the rsultant velocity..Why is the velocit initially attained by car not considered in momentum conservation..????Pllzz do explian sir....Thank you

Admin

2/8/2013 05:28:49 pm

same as above

Prateek

2/9/2013 01:26:20 am

Thanku sir..One last thing i wanna ask is..are the ans of ques 46 and 56 (b) wrong..??
In ques 46 the ans i got is 4.5 cm while in hcv it is 6.1..and in 56 (b) ans is 25l/9 while it is given 2l..??

Prateek

2/9/2013 01:29:42 am

Sorry sir understud the why it is 2l in 56th..please tell about ques 46..Thank You sir..u are a saviour..:)

Admin

2/10/2013 08:40:47 pm

Dear Prateek,
First of all we are sorry for delaying your answer.
In question number 46,
we know that the in the equilibrium condition the spring is stretched by a distance of 1 cm when a mass of 200 g is suspended through it.
Through this we have calculated spring constant (K) = 200 N/m
Now when the mass of 120 g will fall on the block then the spring will stretch further.
we know that all forces acting on the system are conservative. so, we can apply conservation of energy theorem.
u = 3 m/s = velocity of 120 g block just before hitting 2 kg block.
v = 9/8 m/s = velocity of the combination of 2 kg and 120 g block just after the collision.
x = extra deflection of the spring after the collision.
Now at the maximum deflection the block will be at rest which means that whole kinetic energy is converted into potential energy.
Total energy at maximum deflection = Potential energy at maximum extension + potential energy of masses
= (1/2)K(x+0.01)² +P.E1
where P.E1 = Potential energy of masses at maximum extension

Total energy just after collision = Kinetic energy of combination of masses +potential energy of spring + potential energy + Potential energy of masses just before collision
= (1/2)mV² + (1/2)K(0.01)² +P.E2
where m = (.2 + 0.12) = .32 kg
Now P.E2 - P.E1 = mgx
because the masses are coming down by distance x
Now applying energy conservation theorem and equation these energies we get

(1/2)K(x+0.01)² +P.E1 = (1/2)mV² + (1/2)K(0.01)² +P.E2
(1/2)K(x+0.01)² = (1/2)mV² + (1/2)K(0.01)² + P.E2 - P.E1
(1/2)K(x+0.01)² = (1/2)mV² + (1/2)K(0.01)² + mgx
Now solving the above equation we get x = 5.1 cm
and the maximum extension = 5.1 + 1 = 6.1 cm

2/11/2013 04:43:01 am

plz explne example 6

Admin

2/11/2013 08:50:43 pm

Dear RK,
Here we can see that the forces are not acting in the same direction. They are actin at different angles to the horizontal. So, first of all we will try to get the forces along the same direction. So, we will resolve the forces along the horizontal and vertical direction.
Now we will all the components of forces in the x direction and the net force F(x) along x axis = 3.6 N.
Similarly we add all the components of forces along y axis and get the net force F(y) = 1.7 N
Now these forces are acting perpendicular to each other. so, we get resultant force F = √(F(x)² + F(y)²)
Net acceleration = F/M

ruby

2/12/2013 12:35:16 am

in Q 27 we hv given mass=50m bt hw obtn 49m and 48m

Admin

2/12/2013 08:47:30 pm

Dear Ruby,
50 m is the total sum of mass of car, gun, shell and the operator. But when we fire first shot then one shell will be removed form the mass and the mass of one shell = m .remaining mass = 50 m - m = 49 m
Now during the second shot, another shell will be projected away so again there will decrease in mass by m = 49 m -m = 48 m

Prateek

2/12/2013 02:37:17 pm

Sir..I dont know the fundas behind variable mass system involving 'chain problems' etc that involves integration sumtimes..can u suggest me any book..or any link..or any video..i just need to see one or two examples..they will be handful..:)

Admin

2/12/2013 09:08:53 pm

Dear Prateek,
You would see these types of questions in the subjective problems. If you are student of 11th standard then you don't need to worry because in 12 th you will study calculus where you will understand these concepts better. You may see the derivations of different formulas which are involving variable mass so, you may be able to analyze better.

Asnaf Khan

2/15/2013 05:28:18 pm

Sir, would you please explain Q. 19 ... i could'nt get the part where you find time difference.

Admin

2/15/2013 09:05:56 pm

Dear Asnaf,
When the man slips from the roof then its initial velocity is zero.
u = 0
Time taken by it to cover H distance = t1
s = ut +(1/2)at²
where s = H
a = -g
u = 0
we find t1

Again time taken by man to travel (H-h) distance = t2
Again using s = ut +(1/2)at²
s = H-h
a = -g
we find t2.
Time taken by man to travel h distance = t1 - t2

Rajat

2/22/2013 03:25:09 am

In page number 152 question 9 there is a block of mass m on top of block of mass 10m. We have to find the distance moved by bigger block at instant when smaller block reaches the ground. I cannot understand how it is solved in the book...is there another method also so i can understand it? Like taking accelerations , conserving momentum and all..

Admin

2/22/2013 09:56:29 pm

The doubt solving service for HC verma has been disabled for few days. We are very happy with the faith the students have shown in our service. We are hiring more and more teachers and upgrading our features like audio video techniques to solve your doubts.

najeeb mannan

2/24/2013 01:36:21 pm

sir, can you explain elastic and inelastic collisions, please?

palki

3/1/2013 01:51:15 am

i want the worked out problems from work power and energy of hc verma

theyatin

4/6/2013 12:39:58 am

dear palki. . .
as the worked out problems are in book you can ask the question you dont understand .

arjun

5/28/2013 12:41:57 am

sir pls explain q no 43 &
46 (how energy conservation is applied in this question)

arjun

5/30/2013 06:00:48 pm

sir thank u to support a lot.
sir pls ans.. as soon as possible

theyatin

5/30/2013 06:51:33 pm

dear,
dear as the particle after first bounce have same horizontal velocity but vertical velocity has changed and this time let it be
v=e u sin theta
thus now we can calculate all the required values through putting e*u*sin theta in the place of u*sin theta in standard equations

ram

5/30/2013 06:29:27 pm

why in que no 42 the momentum conservation is applied (as friction is present as external force)

pls explain ques no 43 &47

theyatin

5/30/2013 11:32:42 pm

dear,
the change in momentum is due to friction so friction is included in this concept of conservation energy.
you dont need to think about friction while you balance momentum.

theyatin

5/30/2013 11:35:12 pm

dear ram,
dear as the particle after first bounce have same horizontal velocity but vertical velocity has changed and this time let it be
v=e u sin theta
thus now we can calculate all the required values through putting e*u*sin theta in the place of u*sin theta in standard equations

theyatin

5/30/2013 11:43:53 pm

dear,
as conservation of momentum
momentum of bullet= momentum of pendulum +bullet after sollision
mu=(M+m)v
conservation of energy
first energy of pendulum- kinetic energy of bullet=kinetic energy - potential energy of (pendulum +bullet)

mvs saketh

6/13/2013 08:38:35 pm

Admin,, in question no,. 13.. Mr Verma and Mr Mathur are part of the system,, so if they apply any force on boat which is also a part of the system,, the cm position should not change as they are internal forces,, then how did it change ???

Admin

6/13/2013 10:21:15 pm

Dear saketh,
Yatin sir is not well. Please wait 1-2 days.

Mvs saketh

6/13/2013 11:31:17 pm

i got the answer myself,, thankyou

Vaibhav Arya

7/12/2013 01:34:18 pm

sir, pls explain q no 49 & 50

theyatin

7/12/2013 04:31:41 pm

dear,
49. Mass of the two blocks are m1 , m2.
Initially the spring is stretched by x0
Spring constant K.
For the blocks to come to rest again,
Let the distance traveled by m1 & m2
Be x1 and x2 towards right and left respectively.
As o external force acts in horizontal direction,
m1x1 = m2x2 …(1)
Again, the energy would be conserved in the spring.
kinetic energy of spring suppressed = kinetic energy of system about its mean position
(1/2) k × x^2= (1/2) k (x1 + x2 – x0)^2
thus xo = x1 + x2 – xo
x1 + x2 = 2x0 …(2)
also x1=(2*m2/m1+m2)
similarly x2=(2*m1/m1+m2)

theyatin

7/12/2013 04:38:30 pm

dear,
a)velocity of center of mass; =m2*vo + m1* 0/ m1+m2 =m2vo/m1+m2
b) The spring will attain maximum elongation when both velocity of two blocks will attain the velocity of center of mass.
c)
change of kinetic energy= potential energy stored
that is =1/2 k X^2
and change in kinetic energy is difference of kinetic energy of m2 and kinetic energy of center of mass.
that is 1/2 * m2 *vo - 12*(m1+m2) * velocity of center of mass from a)
hence keeping these two equal and solving w can get the answer

Karthik

7/24/2013 01:46:34 am

Respected sir,
Can you explain 3rd question in chapter 9

theyatin

7/25/2013 01:00:11 am

dear,
it is formula of change in position of center of mass
= M1y1+M2Y2 / (M1+M2)
we need to find Y2 hence we can calculate.

Akhil Krishna

7/26/2013 01:58:49 am

Please explain exercise question no. 29 !!

theyatin

7/31/2013 02:21:00 am

dear,
A small block of mass m which is started with a velocity V on the horizontal part of the bigger block of mass M placed on a horizontal floor.
Since the small body of mass m is started with a velocity V in the horizontal direction, so the total initial momentum at the initial position in the horizontal direction will remain same as the total final momentum at the point A on the bigger block in the horizontal direction.

in short the otion of small block has been transferred to both blocks and have speed v'
so now mv+ 0 = (m+M)v'
hence v'= mv/ (m+M)

Akhil Krishna

7/26/2013 02:22:10 am

questn no 38 too !!

Thanks

theyatin

7/31/2013 02:23:52 am

dear,
which part you have problem in?

himani

7/30/2013 12:50:51 pm

sir plz explain ques 45 chapter 9

Abhay Verma

8/1/2013 04:05:28 am

sir please explain the Q.6 , Q.17 of short anwers questions of Chapter 9 Centre of Mass,Linear momentum,collison

theyatin

8/3/2013 03:16:04 am

dear,
Q6
when car accelerates the ob lean backward to retain its rest position.
but the tension in the string pulls the bob forward in direction of the motion of car.
thus motion of the bob as seen from the ground is increasing in momentum.
no its not violation of linear momentum law as there is an external force applied on the bob.

theyatin

8/3/2013 03:17:48 am

dear,
Q17
the person sitting on right side of the figure is likely to fall
because of moment of inertia of the body of person his upper body is still in rest but lower body that is in contact with seat due to friction may move forward (right) as center of mass may be at the center of body the man may fall towards left.

Rachana

7/12/2017 09:08:39 am

Since, the car is accelerating the change in momentum would not be constant.And we can't conserve momentum.Here,the external force is the pseudo force.

Rana

8/21/2013 02:57:41 am

hello..
in the 60 th problem..
we know that the component of a vector in the direction perpendicular to it is a null vector..
how could it be gsin0cos0 (0=theta)
please advise

theyatin

8/23/2013 03:27:28 am

dear,
as acceleration due to inclined plane is
a*sin theta
then horizontal component of acceleration would acceleration*cos theta
thus
a*sin theta * cos theta

ishani

8/30/2013 02:28:54 am

in question no. 2 why thita is taken as origin pls explain

theyatin

8/30/2013 03:08:29 am

dear
its typing mistake
its Q not theta

ishani

8/31/2013 01:37:43 am

why in question no 4 x1 is taken as R?

theyatin

9/1/2013 01:11:46 am

dear,
your question is not clear.

ISHANI

9/1/2013 02:33:27 am

SIR , WHY IN QUESTION NO4 (x1) is taken as R?

theyatin

9/4/2013 04:02:36 am

dear,
as we need to locate center of mass
thus small disc has center of mass which has coordinates as (R,0)
thus x1=R and y1=0

Shraddha Sonkar

9/11/2013 01:22:52 am

Question number 64(page-164).
If mass of sphere is equal to the mass of small particle then they should have exchanged their velocity after collision, the bigger sphere would start moving with velocity "v" and the smaller particle would come to rest. Please explain why this is not so.

theyatin

9/11/2013 02:42:58 am

dear,
as the velocity of particle is not v but v cos theta
as the case you are assuming would have been possible if they would collide so as there center of mass are in same line then it was possible.
but in this case the velocity of particle is calculate w.r.t. center of mass of sphere.

THE STUDIOUS

9/11/2013 10:58:48 am

please explain question number 56 of the chapter centre of mass,linear momentum and collision in HC verma.

theyatin

9/14/2013 03:29:20 am

dear,
Let the velocity of m reaching at lower end = V1
From work energy principle.
(1/2) × m × V1^2– (1/2) × m × 0^2= mg ℓ
 v1 = underroot (2gh)
Similarly velocity of heavy block will be v2 =under root (2gh)
 v1 = V2 = u(say)
Let the final velocity of m and 2m v1 and v2 respectively.
From law of conservation of momentum we can now obtain
v2
now we can calculate v1-v2

the iit man

9/17/2013 10:08:53 pm

sir in your 54 ques you have done a major mistake in the problem solving part
sir please pay atteintion ti the signs

theyatin

9/22/2013 02:08:19 am

dear,
we regret any mistake we will soon quantize it.

Anurag Mittal

9/20/2013 03:25:37 am

i am unable to understand q-11
how to do it if we dont know how to calculate the centre of mass of a semicircular wire using integration

theyatin

9/22/2013 02:22:44 am

dear,
Ycm= difference in volume of circles/ difference in area.
as volume would be half as of whole disk
area is half of the whole disk

Anurag Mittal

9/20/2013 03:29:38 am

in hcv q-3 answer behind is given 22/35 whereas ur answer is 11/35 which one is correct

thyeatin

9/22/2013 02:24:49 am

dear,
answer is 22/35
it mistakenly written 11/35
as 44/70=22/35

raju

9/27/2013 08:05:55 pm

sir, i can't understand the 56th problem of center of mass??

theyatin

9/29/2013 04:26:48 am

dear,
as change difference in force may be F1-F2
that is m1a-(-m2a) or (m1+m2)a=F1-F2
that is a=F1-F2/m1+m2\
that is same in both cases just with difference of signs
net force on m1=F1-m1a (force applied - force acting on m1)
thus we can find it in both cases

now wrok done is
net force on 1*x1 + net force on 2*x2

satyam

10/7/2013 02:29:25 am

sir can u please give me tips to improve my physics
so that i can crack IIT

theyatin

10/8/2013 05:11:41 pm

dear,
just give more stress on basics and practise numerical.
there are thousends of students preparing for iit's
only practice and hard work will make a difference

Harshit Aggarwal

10/16/2013 02:36:21 am

how we have to find the distance of mass of semicircular plate from its centre.

theyatin

10/16/2013 03:37:57 am

dear,
are you talking about center of mass?

Harshit aggarwal

10/16/2013 10:54:15 pm

yes I am talking about center of mass

heyatin link

10/23/2013 01:48:54 am

dear,
as the center of mass is average of mas*distance from all the points in the object.
i have added a link you can study different ways to calculate center of mass of semi circular disk

Abhay Verma

10/16/2013 05:30:20 am

sir please explain question 60

Aymanzoor

10/19/2013 08:09:04 pm

Sir
isn't change in K.E.= final KE - initial KE
Then why in q50 we take we take it as
initial KE-final KE
as the initial vel is only with m1 and is equal to v0 and the finale ke is possessed by the whole system with vel of the centre of mass

theyatin

10/23/2013 02:31:49 am

dear,
as initial kinetic energy is in direction of motion and final is against so it is negative.

Aymanzoor

10/26/2013 12:04:16 am

but how can KE be negative
KE=1/2mv^2
whether v is positive or negative KE will always be positive

theyatin

10/26/2013 02:19:05 am

dear,
its not negative its in opposite direction. . .

Aymanzoor

10/20/2013 02:54:58 pm

Sir
q55 has been done wrongly
s comes out to be 0.24 and not 0.196
now why should i use the last step involving cos30 which i didnt understand why we used in the first place
so is the answer 0.24 sufficient
if not could you please explain why did we use the last cos step

theyatin

10/26/2013 02:40:23 am

dear,
time t=0.24 and s=0.196

lateral distance - perpendicular distance.
here perpendicular distance is dcos theta

Aymanzoor

10/20/2013 05:00:00 pm

sir the answer given in q58 pt b is written as 25m in verma
but both my and the solutions method gets it 12.5m

theyatin

10/26/2013 02:47:14 am

dear,
if your method is right and calculations are correct then no need to worry about answers all you need is to practice numerical.

divyam bhaskar

11/11/2013 11:31:44 am

Sir plz explain q.no. 19

theyatin

11/18/2013 09:56:28 pm

dear,
when the boy is in air if he throws a bag then tee is no ground to apply force on boy hence momentum would be conserved
the moment of bag thrown and boy would be equal.
so mv = MV
now we need to calculate time of flight
difference between time of flight is the time in which boy lands in water.
so V=x / t1-t2
putting it in mv=MV
we have velocity of the bag
As there is no external force in horizontal direction, the x-coordinate of CM will remain at that position.
as gravity acts in vertical direction.
so momentum of c.m. is zero hence
Mx +m x1/ (M+m) =0
hence we can find x1

Arpit

11/18/2013 09:21:25 am

sir can plz explain why in q.no.43 velocity of approach is taken as v whreas it is u

Arpit

11/18/2013 09:22:54 am

i am talking about the lesson centre of mass

theyatin

11/18/2013 10:00:32 pm

dear,
it is just to denote difference between both terms
you can take it as u1

Arpit

11/23/2013 08:43:08 am

oh!!!thank you sir

saloni kumari

11/18/2013 03:34:49 pm

sir plz tell me question no 56

theyatin

11/18/2013 10:02:41 pm

dear,
are you asking to explain question??
or you want answer??

ankit rai

11/19/2013 03:24:48 pm

thanks sir

sofi link

11/19/2013 03:48:42 pm

Sir, pls explain d Qno.3 of chptr(9) whch equation should be used in dis Qustn??

theyatin

11/19/2013 04:47:11 pm

dear,
the eqn is
m1d1+m2d2+m3d3.....+m7d7/m1+m2....+m7

ankit rai

11/19/2013 09:36:53 pm

please explain solution of the question no.14 chapter-9 exercise?

theyatin

11/22/2013 02:10:38 am

dear,
Initially the monkey & balloon are at rest.
So the CM is at ‘P’
When the monkey descends through a distance ‘L’
The CM will shift
this implies that the center of mass would also lower as monkey goes down.
m*L+M*0 / m+M is the position of center of mass whee balloon is still at rest and monkey has descended down.

ankit rai

11/19/2013 09:43:49 pm

sir ,please explain the answer of question -6 MCQ, CHAPTER -9

theyatin

11/22/2013 02:05:54 am

dear,
which objective please specify ??

arpit link

11/22/2013 02:09:19 am

sir why we took distance from centre of mass tobe 4r/3pie in Q11?

shivam

11/28/2013 09:10:01 am

please explain q.no. 41 &42

theyatin

11/29/2013 01:27:03 am

dear,
Q41,
Mass of each block MA and MB = 2kg.
Initial velocity of the 1st block, (V) = 1m/s
VA = 1 m/s, VB = 0m/s
Spring constant of the spring = 100 N/m.
The block A strikes the spring with a velocity 1m/s/
After the collision, it’s velocity decreases continuously and at a instant the whole system (Block A + the
compound spring + Block B) move together with a common velocity.
Let that velocity be V.
initial energy of A+ initial energy of B= final energy of A +final energy of B
as initial energy of B is zero because it is at rest
final energy of B is due to spring where spring constant is given
on solving this we have
1= 2v^2 + 50x^2
now conserving momentum
initial momentum of B is zero and final velocity of B is equal to A as system is moving with same velocity.
so we get 2+0 = (2+2)*v
so we have v=1/2
putting value of v we can have x.

theyatin

11/29/2013 01:47:31 am

dear,
Q42
first we need to conserve momentum
initial momentum of block is zero
mv1 + Mu2 = mv + Mv'
hence form here we can find v'
now as we know work done= change in momentum.
thus 1/2 *m(0^2-v'^2)
also work done= f*d= uF*d
comparing both equations. . .
we can have u

shubh

11/28/2013 02:46:14 pm

In finding resultant acc. Of small mass m w.r.t wedge. Why a2 is in left dir. Can we solve it using enegy conservation law but l m unable to get relation beteen velocities of mass and wedge !!!

theyatin

11/29/2013 02:04:59 am

dear,
when the mass slide down
the bock of mass M would also move forward (left direction)
as the vertical component of weight of small block would push big block forward.

shubh

11/28/2013 02:50:15 pm

Sorry I forgot to mention Q no. Its 60

shivam

11/29/2013 11:06:59 pm

sir
if 1000kg of block with u=5m/s collide with spring mounted horizontally &friction b/w surface is 0.5 then what is max.compression in spring ?
please explain.

theyatin

11/30/2013 02:04:26 am

dear,
let compression is x
then work done=change in kinetic energy
while work done=uF*x
comparing both
1/2*1000*5^2=0.5*1000*10*x
5/2=x

sumaley link

12/5/2013 02:12:29 am

qustion 15 nhi samaj aa rha hai

shubhu

12/5/2013 12:54:16 pm

Just use the relation of K.E and linear momentum i.e K=p²/2m and just equate as required!!!

Neeraj

12/7/2013 12:15:33 am

In HCV the answer for q46 is given as 6.1cm but here it is given as 4.5cm which one is correct?

anushya

12/7/2013 07:08:57 pm

can u please explain q10 of centre of mass.
why will the ice melt and aquire a spherical shape?
if the shape of the boday changes, doesnt the cetre of mass change? in this sum, why is the shift in com 0?

varun

12/7/2013 07:27:31 pm

even i havent understood q10,sir can you please explain?
why is it necessarily spherical ?

shubhu

12/11/2013 12:51:39 am

Due to absence of any net attraction force on liquid it will acquire spherical shape due to surface tension. And since there is no external force c.m will not chage

Shreyas

12/11/2013 12:22:28 am

sir....in question 53........how is it possible that the pillow and man reach reach at the same initial position.......since there is no gravity ....how can the man come down again?

theyatin

12/22/2013 12:48:04 am

dear,
as the external force on the system is zero
so the initial action is equal to final action. . . or in other words reaction.

Shreyas

12/22/2013 02:06:06 pm

but sir theres not even internal forces between them which will bring them together......i dont think that the man will come down.......and also theres no specification that the speed of pillow is relative to the man.............
i have got the ans much more precise with a different approach..
i think that the speed of pillow is with respect to ground....So
Vman=0.8 ft/sec and Vpillow=8ft/sec.......
So after 2 secs.......pillow reaches the initial position and the man travelled a distance of 1.6 ft...upward
(the pillow reaches mans head).....
therefore Spillow - Sman=1.6
=> 8t - 0.8t =1.6
=> t=1.6 / 7.2 => t=0.2222 sec
therefore pillow reaches mans head at 0.22 secs..... after pillow reaches the initial position........
therefore TOTAL TIME TAKEN= 2+ 0.22=2.22(much more precise than the above specified ANS)
So, i find this approach to be much more credible than the given one...........Send in your views..

Shreyas

12/22/2013 02:07:10 pm

but sir there's not even internal forces between them which will bring them together......i dont think that the man will come down.......and also there's no specification that the speed of pillow is relative to the man.............
i have got the ans much more precise with a different approach..
i think that the speed of pillow is with respect to ground....So
Vman=0.8 ft/sec and Vpillow=8ft/sec.......
So after 2 secs.......pillow reaches the initial position and the man travelled a distance of 1.6 ft...upward
(the pillow reaches mans head).....
therefore Spillow - Sman=1.6
=> 8t - 0.8t =1.6
=> t=1.6 / 7.2 => t=0.2222 sec
therefore pillow reaches mans head at 0.22 secs..... after pillow reaches the initial position........
therefore TOTAL TIME TAKEN= 2+ 0.22=2.22(much more precise than the above specified ANS)
So, i find this approach to be much more credible than the given one...........Send in your views..

Shreyas

12/22/2013 02:07:51 pm

theyatin

12/24/2013 11:13:50 pm

dear,
from where did you get velocity of man is 0.8ft/sec??

Shreyas

12/25/2013 12:33:59 am

sir .....by applying momentum conservation of (man +pillow)system:

0 = M*Vman-m*Vpillow

=>Vman=(m*Vpillow)/M = (5*8)/50 = 0.8 m/sec........

Your views are highly appreciated...........

Shreyas

12/25/2013 12:36:57 am

not m/sec......its ft/sec.......
doesnt it look more appropraite to use this approach sir.....

reply soon sir

theyatin

12/27/2013 03:34:24 am

dear,
yes your answer is very much precise good job.
first pillow takes time to reach its original position
then meanwhile the distance covered by man must be covered by pillow too.

Pratik

12/24/2013 01:22:25 pm

Sir in example 17 & 24 why we don't consider gravitational force as external force & in presensence of an external force how can we conserve momentum

theyatin

12/24/2013 10:48:40 pm

dear,
if we conserve momentum of two bodies in dynamics
we only consider two objects w.r.t to each other no matter what is cause of their motion.
we only consider velocities. . .

Shreyas

12/25/2013 01:06:08 am

in example 17........the (man+earth) is the system.......therefore gravitational force acts as an internal force between them...

similarly in example 24........consider (hailstones+roof+earth) as the system .....gravitational force then acts as an internal force between the particles......hope it clears ur doubt

Krishna

9/29/2016 01:31:57 pm

Hi,
I understood ur doubt very well. Though replying after 3 yrs, it might help other learners having this same doubt. Read the article here: http://www.jeequery.com/87/

Pratik

12/25/2013 11:02:15 pm

But we take hailstone +earth + roof as a system don't we have to calculate earth in momentum conservation as the earth has a very large mass??

theyatin

12/27/2013 03:01:33 am

dear roof is in contact with earth so eventually it is part of earth
and in case you are balancing momentum you only need to balance it between system of considered objects ignore all others.

Shreyas

12/26/2013 12:14:06 am

observer is on the earth =>earth is at rest with respect to observer

theyatin

12/27/2013 02:48:55 am

dear,
yes both are at rest w.r.t each other as both are in contact both move together.

aditya sahu

12/31/2013 04:05:31 pm

Please help me I don't know how to do hcv question number 59 from rotational motion page number 199

theyatin

12/31/2013 10:53:38 pm

dear,
from work energy principle
change in kinetic energy=work done
Let velocity of 2kg block on reaching the 4kg block before collision =u1.
we can calculate u1
and if collision is elastic
we can conserve momentum of the system
from this we can find final velocities
now we need to use work energy theorem for each block
m=2kg
& M=4kg
and calculate distance in each case for work done.

Shreyas

12/31/2013 11:58:30 pm

a)change in linear momentum(of the rod) = impulse(Chapter 9)
=> mv=Ft
v =Ft/m

b)change in angular momentum(about CM) =angular impulse(J)
=> m(L^2/12) (w) = F(L/2) * t
=> w = 6Ft/mL

c) KE = KE(rotational)+ KE(translational)
= (3F^2t^2)/2m + (F^2t^2)/2m
= (2F^2t^2)/m

d) L = I w (I and w are same as earlier)
=FLT/2

vishwa

1/1/2014 07:16:05 pm

the explanation you gave in ques 23 is not understood to me..please explain in another way. i interpreted the solution that as the system is isolated the mass lossed by astrounauts will remain in spaceship so there will be no change og velocity. Is it correct?

theyatin

1/3/2014 01:30:20 am

dear,
everything what happens is inside the spaceship no mas actually leaves the ship
as total mass is still conserved thus velocity is.

Shreyas

1/1/2014 08:15:42 pm

yes it is correct........i too had faced problems in this particular question......the mass lost by the astronauts are converted to some equivalent form of energy(by doing exercise-the energy required to do the exercise is produced by the fats stored in astronauts-making them to loss mass) .....so this energy can be again converted to form equivalent amount of mass....E=mc^2 .........and i m also a student....sir if m wrong anywhere please point it out....listening to others doubts really helps a lot

theyatin

1/3/2014 01:21:15 am

dear shreyas i am glad that you people are helping each other but you need to answer the question in the reply option otherwise person asking question may not know that you have answered.

Mohak

1/13/2014 03:59:40 pm

Sir in que 57 why we take dx/dt as dv and what is the physical significant of this?

Shreyas

1/13/2014 08:49:30 pm

v = dx/ dt

velocity is rate of change of displacement

theyatin

1/14/2014 01:49:01 pm

dear derivative of distance w.r.t to time means change of distance w.r.t. time that is velocity or motion

Mohak

1/14/2014 02:37:46 pm

But sir we have considered that dx is the elemental length of mass dm then why we consider it as a distance...

theyatin

1/20/2014 01:42:35 pm

dear,
dx is infinitesimal small length it is not distance
but when we take derivative it w.r.t. to an instant of time we get velocity.
it is not simply change in distance over change in time but it is derivative of length w.r.t time.

Mohak

1/17/2014 04:48:41 pm

Sir I haven't got my answer so far...sorry!!

Jay

1/20/2014 12:15:03 am

Sir I got same question as mohak

theyatin

1/20/2014 01:43:12 pm

dear,
dear,
dx is infinitesimal small length it is not distance
but when we take derivative it w.r.t. to an instant of time we get velocity.
it is not simply change in distance over change in time but it is derivative of length w.r.t time.

Mohak

1/21/2014 10:23:13 pm

Sir how can we solve que 60 using energy conservation?

theyatin

1/23/2014 01:04:09 am

dear, please show me how are you proposing to attempt this question?? this way

Shreyas

1/23/2014 01:23:28 am

sir in SHM page no.253 ......Question no.19,20......plz explain how to get the forces exerted by spring A & B ...especially in Q.20...plz help

theyatin

2/19/2014 01:08:38 am

dear,
please ask ur questions in s.h.m. chapter

Mohak

1/23/2014 12:47:43 pm

mgh= 1/2mv^2 + 1/2MV^2
As no external force in horizontal direction when block reach the bottom,
[email protected]=MV
solving both I'm getting,
V=[2m^2ghcos^[email protected]/M(M+mcos^[email protected])]^1/2

theyatin

2/19/2014 12:57:03 am

but dear it is not required answer.

abhinav

2/18/2014 09:42:08 pm

sir in q57 we took dx as the smallest possible elemental length..then when we found out the opposing force by the ground for its momentum to be zero..but isnt it for the small element only..
i mean the elements below would also attain some momentum..
as you can see im reallly confused could you please explain it very CLEARLY..!!

theyatin

2/19/2014 01:05:20 am

dear,
i cant understand your question
we consider force for small length of element then calculating for whole.
where did you find force is zero.

abhinav

2/19/2014 04:00:49 pm

sir mujhe ye smajh nhi aaya ki pehle unhone dx length ka momentum calculate kra fir jab rate of change of momentum nikala tab wo dx/dt kiski velocity represent kr rha hai...

sachin jha

3/6/2014 11:23:49 am

sir what is thermodynamics with its application and its proof?

Admin

3/6/2014 03:58:42 pm

Please wait. Yatin sir will resume work within 2-3 days.

Aditya

3/7/2014 02:41:53 am

In Q 43
why is e=usin<>/v
velocity of approach = u
velocity of sep.=v
e=u/v
v=eu
vcos<alpha>=ucos<> --->alpha=angle of second trajectory
is my expression wrong i couldnt solve using my mehod

theyatin

3/19/2014 01:37:35 am

dear,
v cos alpha= u cos alpha????
how did you get it??
i dont think its appropriate. please elabourate

swadhin agrawal link

3/8/2014 09:16:40 pm

sir I am unable to know step m1x1=m2x2 in question no. 49 0f centre of mass.

theyatin

3/19/2014 01:28:09 am

dear,
you need to understand that in spring system m is inversly proposrtional to distance thus m1 inversly proportional to x1 so as m1x1=constant
and m2 is inversly proportional to x2 so as m2x2=constant
thus in this system m1x1=m2x2

Swadhin Agrawal link

3/27/2014 01:07:04 am

Sir,but I want to know that why in a spring system this mass is inversely proportional to the distance? Sir, is it by the application of centre of mass in a system?

theyatin

3/28/2014 11:33:09 pm

dear,
as foce of spring is directly proportional to square of distance from mean
thus similarly this statement has been made.

swadhin agrawal link

3/31/2014 06:49:33 pm

Sir, I am confused, I could not get it.

theyatin

4/1/2014 04:35:42 am

dear,
force is constant thus only fctors that may affect the motion are mass and distance from mean.
as force is due to mass and distance
to keep force constant either mass may be greater so distance must be lesser
or mass may be lesser and distance must be greater.

swadhin agrawal link

4/1/2014 03:39:27 pm

yes sir, I get it. Thankyou sir.

Swadhin Agrawal link

4/1/2014 03:43:05 pm

sir I have posted my doubts in next chapter rotational mechanics. please sir check them out.

pavandeep kaur

4/10/2014 10:50:20 pm

can u pls explain the 49 th sum of collision

4/21/2014 05:18:11 pm

y du v need 2 tke d sum of velocities in 27q plz exp in detail!!!!

xwedeerf

4/21/2014 10:59:12 pm

suggest me sum other method 2 solve d 27q plz!!!!!!

niti n

4/21/2014 11:01:06 pm

suggest me sum other method 2 solve d 27q plz!!!!!!
AT D EARLIEST!!!!!!!!!

swadhin agrawal link

4/22/2014 05:14:50 pm

Hello,
apply the equation
w=wo+cxt
Cx means alpha

niti n

4/24/2014 03:11:27 pm

hi,swadhin....thx but cud u explain dat further???
wat is w=wo+cxt........frm wich chap. is it ....plz give d pg no. nd d chap. name frm where u 've derived dis eq.(w=wo+cxt)....plzreply 2 it at d earliest!!!!!!!!
andI THOUROUGHLY APPRECIATE UR HELP!!1

swadhin agrawal link

4/24/2014 03:32:32 pm

Hey it is the eeuation of motion in circular path see chapter circular motion it is there.

niti n

4/24/2014 03:14:16 pm

has 'ny1 got d alternate method 2 solve d 49Q....plz reply!!!!
sir can u suggest me sum 'ther method 2 solve it??????????????

ravi kumar

5/29/2014 02:01:04 am

sir explain about collision

mirjaik

5/31/2014 12:29:33 am

an instance of one moving object or person striking violently against another

admin

6/22/2014 01:47:20 pm

Dear Students,
Our Teacher is unavailable for few days. We are sorry for this inconvenience. Our teacher will give your answers soon.

shubham

7/6/2014 02:36:17 pm

i didnt get chapter 9 exercise que nos 35,51,48,60,61 please replay fast...

admin link

8/8/2014 01:45:12 am

Dear shubham
if u didn't uderstand the question then u donot know the basics ! u r too poor in physics ! how the hell do u think ill be able to give sol for those many questions ! go to other website ! chal chal !

admin link

8/8/2014 01:42:12 am

Sumedha Ghoshal

8/9/2014 12:51:20 am

sir cld u xplain me ques 25 and 33.

Sumedha Ghoshal

8/9/2014 12:58:10 am

sir pleas explain example 19 of chapter 9 the line "As the centre of mass is at rest in this frame the blocks move in opposite directions and come to intantaneous rest at some instant.The extension of the spring will be max at this instant." PLEASE do reply .

Sumedha Ghoshal

8/9/2014 03:28:12 am

the sums having two bodies connected by a spring are creating problem for me.i couldnt understd ques 49 soln. PLEASE HELP !!!!

Sumedha Ghoshal

8/14/2014 09:08:15 pm

u hav promised to deliver answer within 24 hrs but i have waited for
more than 6 days now u ppl dont understd value of time for a student

nvp aditya link

9/1/2014 01:51:13 am

this is so good thank you for the solutions this actually increases my knowledge

Vivek Om

9/11/2014 02:28:20 am

this is to all if anyone knows solution of 4th que. Pls leave a reply..

Sashank

9/11/2014 07:50:16 pm

Sir thanks a lot. This has helped me a lot. I have a doubt in 60th question. Shouldn't acceleration in horizontal direction be 0. You guys seemed to have conserved acceleration rather than velocity(or maybe I understood it wrong, I have not quite got how you got the acceleration of the big block) if we take the horizontal plane at an angle Alfa to the incline, and the vertical plane 90- Alfa the answer will vary greatly. Because then horizontal momentum should be zero. So velocity of big block w.r.t small block horizontal velocity is found. And so on.

Thanks a lot for this help

Anand gupta

9/21/2014 11:05:42 pm

sir in que no... 46 ...how to find the spring constant ....

Om Patil

9/26/2014 10:00:27 pm

I am unable to understand q61 of exercise .PLEASE HELP ME.

Harshit

9/29/2014 01:27:28 am

sir please tell me how to deal with the problems...i am not knowing how to start.......please help me..

10/3/2014 11:25:45 pm

abhishek

10/3/2014 11:27:44 pm

sir plz explain q 36 of hc verma collision

saraswati

10/6/2014 01:31:11 am

sir can you explain how did you take sin52 x 0.96 A as x1 in 2 problem

barsha

10/11/2014 03:44:53 pm

pls explain me the difference between the question 4 and 5

In question 4 there are two discs placed one over another and in question 5 there is one disc of radius 2R from which a disc of radius R is cut out.

10/16/2014 06:06:20 am

Rupak

10/14/2014 06:31:16 am

sir can you explain problem no. 44 a bit?

shivam

11/17/2014 02:38:50 am

sir how we can identify where the centre of mass lying between the combination of system for example ques6 of objective1 in hc verma

Shivam Anand

12/10/2014 11:25:08 pm

Sir, please help me out with q.no.-52 of chapter-9..

saurabh

12/25/2014 02:00:27 pm

chapter 9 question 50 b

viraj

1/6/2015 10:23:50 am

Sir can u tell me how do the trajectory equation in que no. 43 is plz explain it..

tanvee

3/24/2015 12:58:31 pm

a sphere with velocity v strikes a wall moving towards it with velocity u. if the mass of walll is infinetly large then work done by the wall during perfect elastic collision will be ?

Zlatan Ibrahimovic

4/21/2015 03:08:56 pm

The ans. for the third que is 22L/35 but the solution says 11L/35..?

prabhat mishra

5/18/2015 08:37:48 pm

Sir,,please explain question no 60,61 from centre of mass

sir question no 59 hcv

8/9/2015 05:27:13 am

nikhil

9/27/2015 09:53:58 pm

u can ask me on my email id. i will send u the pic of soln

saurabh

9/24/2015 09:16:28 pm

Hints for the q-13 and q-14 of h.c verma...conservation of linear momentum.....

nikhil jha

9/27/2015 09:43:35 pm

ques 61.. why cant we use equation of a projectile(kinematics) to find the maximum height reached by the block taking our reference point as the top of wedge of mass M. then we can add the answer to the height h to get the final answer.
is this method right ?

sameer maurya

10/9/2015 05:19:34 am

sir i could understand Q..49 what is x in it. please explain me.

bhoopesh

10/9/2015 07:26:49 am

Please explain friction 21 question exercise

pratiksha link

10/20/2015 06:39:06 am

Sir,can u explain me Q 11 of exercise.......

kanjal

10/22/2015 05:42:26 pm

can u please explain question 13 to me

caersar

11/3/2015 10:18:31 pm

good initially but tons of misprints in many questions... horrible experience very frustrating.. sorry but take more care while publishing such important solutions.. so many solutions aren't clear due to wrong printing..students have a tough time deciphering what the solutions actually mean if you use the wrong terminologies like foe e.g.: u1 instead of v2, .6 in one line and 6 in the otheretc.. please look into this matter.

Jai link

11/17/2015 06:56:50 am

hey in hc verma book the ans for qs.46 is 6.1 cm

vaishnavi

12/13/2015 06:07:28 am

aren't question 4 and 5 same?? then why is the answer different??

dipssy

12/25/2015 09:37:29 pm

perfect solutions

Riya

12/28/2015 08:30:24 am

Sir kindly explain me the prlb no.24 of hcv chapter- linear momentum

innocent

3/8/2016 07:44:20 pm

A ring and a disc having the mass rolls without slipping with same linear velocity.if the kinetic energy of the ring is 8J.calculate the kinetic energy of the disc

Manish moryani

3/19/2016 04:01:15 am

Four.
Mr^2=8
So,Mr^2/2=4

sadik ali

7/15/2016 07:38:00 pm

sir plece solution for q2

Agniva Roy

7/22/2016 08:02:12 am

Question no 52 of centre of mass impulse momentum.How can we say that acceleration will be F1-F2/m1+m2.How do we come to this conclusion?
question number 51
why will the elongation be most when common velocity will be velocity of cm?Why will it not be any other case(example when velocity of both is zero)?

jones

8/23/2016 09:36:53 pm

i dont understand 11 ques of this chap
how did u get the first eq

rajan

9/2/2016 12:17:58 am

Sir, I facing problem in question for short answer in 9 CH of class11.
Please help me.....

sagar link

9/9/2016 07:36:09 am

good

Priyamvada

9/12/2016 11:26:18 pm

Can u explain me Q51 that how the equation 3 came???

sweety priya

9/26/2016 09:30:25 am

Q.1of chap.centre of mass short answer type question

Abhishek sawan

10/9/2016 08:02:04 am

Capeter 9 questions no 19 hc verma

Toni

10/16/2016 05:51:55 pm

Ch 9 question no 7 of. HC Verma book

Abhinav

6/13/2017 06:48:20 am

Hcv example no. 17 why the linear momentum is being conserved though there is external acc. 'g' acting downwards in vertical direction. please reply soon

Utkarsh

10/20/2017 10:05:07 pm

Pls explain ques no 56 I am finding some errors in the soln

Anupria

10/6/2018 12:10:52 am

Plz xplain centre of mass solved examples number 13 .how the mass of module is M/6

ARYAN

8/30/2019 08:02:45 pm

PLEASE EXPLAIN THIS TO ME

in question 59 i have found out the accelerations of the two blocks to be 2 m/s^2 in opposite directions
then i have used the concept of relative velocity and relative acceleration to find out the separations the answer DOES NOT MATCH
IT WOULD BE A GREAT HELP if you could solve using relative motion PLEASE

Chapterwise Solutions of HC Verma