Can conservation of momentum be applied here?

[ajaysabarish]

Take the case of Atwood machine, Where the masses of blocks are m and m,now,another ball of mass m strikes one of the blocks with a speed v and sticks to it.

Considering our system as blocks+ball+string connecting the blocks(no pulley),can we apply conservation of momentum?

Please explain why?our system is blocks+ball+string and no pulley.

Assume impulse due to gravity is negligible(since time of collision is small)

edit 1: i have used the word atwood machine to just make the readers understand the situation.

 

[ehild]

The ideal Atwood Machine consists of two objects of mass m1 and m2, connected by an inextensiblemassless string over an ideal massless pulley. [1]

https://en.wikipedia.org/wiki/Atwood_machine#cite_note-1

Without the pull,y it is not an Atwood machine.

 

[ajaysabarish]

Without the pull,y it is not an Atwood machine.

thank you for replying sir,but i just used the word atwood machine to make the readers understand the situation.please reply to my question.

 

[conscience]

Take the case of Atwood machine, Where the masses of blocks are m and m,now,another ball of mass m strikes one of the blocks with a speed v and sticks to it.

Considering our system as blocks+ball+string connecting the blocks(no pulley),can we apply conservation of momentum?

Please explain why?our system is blocks+ball+string and no pulley.

Assume no gravity acts anywhere.

Is the ball moving vertically or horizontally before colliding with the block ?

 

[ajaysabarish]

Is the ball moving vertically or horizontally before colliding with the block ?

thank you for replying sir,sorry for inadequate information,ball strikes it vertically.the entire setup is placed in a vertical plane only,but ignore the gravity.

 

[conscience]

Do you think there are any vertical forces acting on the system (blocks+ball+string) during the collision ?

 

[ajaysabarish]

Do you think there are any vertical forces acting on the system (blocks+ball+string) during the collision ?

i think there will be a force on strings due to pulley.as strings apply a tension force on pulley,pulley also applies a force on system,as pulley is not a part of our system.and even if there is gravity(as in original question),the impulse will be very small since time of collision is small therefore it can be neglected.but in solution it is given that momentum is conserved,but i couldn't understand why.please help me sir.

 

[conscience]

i think there will be a force on strings due to pulley.as strings apply a tension force on pulley,pulley also applies a force on system,as pulley is not a part of our system.and even if there is gravity(as in original question),the impulse will be very small since time of collision is small therefore it can be neglected.but in solution it is given that momentum is conserved,but i couldn't understand why.please help me sir.

Do you think there would be any tension in the strings if there was no gravity ?

 

[ajaysabarish]

Do you think there would be any tension in the strings if there was no gravity ?

yes sir i forgot that,i will edit the question and i am sorry for wasting your time.

 

[ajaysabarish]

Do you think there would be any tension in the strings if there was no gravity ?

now i have changed the question sir,please clear my doubt.

 

[conscience]

i think there will be a force on strings due to pulley.as strings apply a tension force on pulley,pulley also applies a force on system,as pulley is not a part of our system.and even if there is gravity(as in original question),the impulse will be very small since time of collision is small therefore it can be neglected

The force on our system is gravity and the force due to the pulley . Impulse due to gravity can be neglected , but do you think impulse due to force exerted by pulley can be neglected as well ?

Assuming pulley is massless what is the force exerted by the pulley on the system ?

Note : Can you write down the exact question (word by word) you are trying to solve ?

 

[ehild]

thank you for replying sir,but i just used the word atwood machine to make the readers understand the situation.please reply to my question.

If there is no pulley, the momentum would be conserved. But the pulley is there and it exerts force on the system masses+string. Momentum is conserved if there is no external force or it can be ignored.
In what direction will the mass hit move after collision? And what do you think, in what direction will the other mass move?

 

[conscience]

But the pulley is there and it IS part of the system.

I think what the OP is trying to convey is that pulley is present in the setup , but is not considered as part of the system .The system comprises of (blocks+ball+strings) .

 

[ajaysabarish]

The force on our system is gravity and the force due to the pulley . Impulse due to gravity can be neglected , but do you think impulse due to force exerted by pulley can be neglected as well ?

Assuming pulley is massless what is the force exerted by the pulley on the system ?

Note : Can you write down the exact question (word by word) you are trying to solve ?

sure sir,

2 blocks each of mass m are hanging as shown in the figure(https://www.google.co.in/search?q=atwood+machine&biw=1600&bih=775&tbm=isch&imgil=J-26XXHLI8jOsM%253A%253BmhX7Xm1xxTHICM%253Bhttp%25253A%25252F%25252Fphysics.stackexchange.com%25252Fquestions%25252F118917%25252FNewtons-third-law-and-atwood-machines-confusion-about-tension&source=iu&pf=m&fir=J-26XXHLI8jOsM%253A%252CmhX7Xm1xxTHICM%252C_&usg=__u_UL0DkMewZLskgjl63jV063Kkw%3D&ved=0CDIQyjdqFQoTCM_g_ZfRiskCFcSalAodIisASQ&ei=RWREVo-qBsS10gSi1oDIBA#imgrc=J-26XXHLI8jOsM%3A&usg=__u_UL0DkMewZLskgjl63jV063Kkw%3D

another ball of mass m is dropped from a height h,the ball hits one of the block and sticks to it,find the velocity of the other block after collision.

it gives 2 approaches,one is impulse momentum theorem which i understood.the second approach solves the problem using conservation of momentum,but it says momentum is conserved in the system taken(ball+blocks+string),but the string is made horizontal and the initial momentum is mu(if ball strikes the block with speed u) and final momentum is 3mv(if final speed of blocks is v).from conservation of momentum
mu=3mv
v=u/3.
but if strings are placed in vertical system itself,initial momentum is mu and final momentum becomes mv+mv-mv(downwards is taken positive and the other block moves upwards).from conservation of momentum,
mu=mv
u=v.
from impulse momentum theorem and the conservation of momentum of theirs,answer is v=u/3.but according to my approach it is v=u.

but i felt conservation of momentum cannot be applied, as there is a net external force of 2T applied on string(which is a part of our system)due to pulley.

image is attached here:https://www.google.co.in/search?q=atwood+machine&biw=1600&bih=775&tbm=isch&imgil=J-26XXHLI8jOsM%3A%3BmhX7Xm1xxTHICM%3Bhttp%253A%252F%252Fphysics.stackexchange.com%252Fquestions%252F118917%252FNewtons-third-law-and-atwood-machines-confusion-about-tension&source=iu&pf=m&fir=J-26XXHLI8jOsM%3A%2CmhX7Xm1xxTHICM%2C_&usg=__u_UL0DkMewZLskgjl63jV063Kkw=&ved=0CDIQyjdqFQoTCM_g_ZfRiskCFcSalAodIisASQ&ei=RWREVo-qBsS10gSi1oDIBA#imgrc=J-26XXHLI8jOsM:&usg=__u_UL0DkMewZLskgjl63jV063Kkw=

please clear my doubt sir.

 

[ajaysabarish]

If there is no pulley, the momentum would be conserved. But the pulley is there and it exerts force on the system masses+string. Momentum is conserved if there is no external force or it can be ignored.
In what direction will the mass hit move after collision? And what do you think, in what direction will the other mass move?

thank you for replying sir,
image is attached here,
https://www.google.co.in/search?q=atwood+machine&biw=1600&bih=775&tbm=isch&imgil=J-26XXHLI8jOsM%3A%3BmhX7Xm1xxTHICM%3Bhttp%253A%252F%252Fphysics.stackexchange.com%252Fquestions%252F118917%252FNewtons-third-law-and-atwood-machines-confusion-about-tension&source=iu&pf=m&fir=J-26XXHLI8jOsM%3A%2CmhX7Xm1xxTHICM%2C_&usg=__u_UL0DkMewZLskgjl63jV063Kkw=&ved=0CDIQyjdqFQoTCM_g_ZfRiskCFcSalAodIisASQ&ei=RWREVo-qBsS10gSi1oDIBA#imgrc=J-26XXHLI8jOsM:&usg=__u_UL0DkMewZLskgjl63jV063Kkw=

a ball of mass m is dropped from a height h and it strikes one of the block.

 

[ajaysabarish]

I think what the OP is trying to convey is that pulley is present in the setup , but is not considered as part of the system .The system comprises of (blocks+ball+strings) .

yes sir,pulley applies a force on our system but pulley is not considered to be a part of our system(ball+blocks+string)

 

[ehild]

The string is around the pulley. The pulley rotates about a hinge. The hinge is supported by the ceiling.
It is weird saying that the pulley is not part of the system that consists of an Atwood machine. But the support and the ceiling and the whole house is not part of it and the support can exert as big force as needed to keep the centre of the pulley at its place. The support exerts an impulsive force F_i and it is not true, that F_idt can be ignored during the collision time dt.
Because of that impulsive external force, the linear momentum is not conserved.
But there is an other conservation law: Conservation of angular momentum. The external torque is due to gravity alone, and its effect can be ignored during the collision time. Assume the pulley has radius r, and the falling body hits the hanging mass just above its CoM. What is the angular momentum of the falling body, and what is the angular momentum of the system of three masses after the collision? You have to know how the velocities of the masses connected to the pulley are related.

 

[ajaysabarish]

respected sir,
anything can be considered as a system even a single block can be considered as a system.but we have to check whether conservation of momentum can be applied or not.so i have taken ball+blocks+string as our system.

is linear momentum conserved sir?

 

[ajaysabarish]

respected sir,
anything can be considered as a system even a single block can be considered as a system.but we have to check whether conservation of momentum can be applied or not.so i have taken ball+blocks+string as our system.

is linear momentum conserved sir?

this is my question and my doubt.

2 blocks each of mass m are hanging as shown in the figure(https://www.google.co.in/search?q=atwood+machine&biw=1600&bih=775&tbm=isch&imgil=J-26XXHLI8jOsM%253A%253BmhX7Xm1xxTHICM%253Bhttp%25253A%25252F%25252Fphysics.stackexchange.com%25252Fquestions%25252F118917%25252FNewtons-third-law-and-atwood-machines-confusion-about-tension&source=iu&pf=m&fir=J-26XXHLI8jOsM%253A%252CmhX7Xm1xxTHICM%252C_&usg=__u_UL0DkMewZLskgjl63jV063Kkw%3D&ved=0CDIQyjdqFQoTCM_g_ZfRiskCFcSalAodIisASQ&ei=RWREVo-qBsS10gSi1oDIBA#imgrc=J-26XXHLI8jOsM%3A&usg=__u_UL0DkMewZLskgjl63jV063Kkw%3D

another ball of mass m is dropped from a height h,the ball hits one of the block and sticks to it,find the velocity of the other block after collision.

it gives 2 approaches,one is impulse momentum theorem which i understood.the second approach solves the problem using conservation of momentum,but it says momentum is conserved in the system taken(ball+blocks+string),but the string is made horizontal and the initial momentum is mu(if ball strikes the block with speed u) and final momentum is 3mv(if final speed of blocks is v).from conservation of momentum
mu=3mv
v=u/3.
but if strings are placed in vertical system itself,initial momentum is mu and final momentum becomes mv+mv-mv(downwards is taken positive and the other block moves upwards).from conservation of momentum,
mu=mv
u=v.
from impulse momentum theorem and the conservation of momentum of theirs,answer is v=u/3.but according to my approach it is v=u.

but i felt conservation of linear momentum cannot be applied, as there is a net external force of 2T applied on string(which is a part of our system)due to pulley.

image is attached here:https://www.google.co.in/search?q=atwood+machine&biw=1600&bih=775&tbm=isch&imgil=J-26XXHLI8jOsM%3A%3BmhX7Xm1xxTHICM%3Bhttp%253A%252F%252Fphysics.stackexchange.com%252Fquestions%252F118917%252FNewtons-third-law-and-atwood-machines-confusion-about-tension&source=iu&pf=m&fir=J-26XXHLI8jOsM%3A%2CmhX7Xm1xxTHICM%2C_&usg=__u_UL0DkMewZLskgjl63jV063Kkw=&ved=0CDIQyjdqFQoTCM_g_ZfRiskCFcSalAodIisASQ&ei=RWREVo-qBsS10gSi1oDIBA#imgrc=J-26XXHLI8jOsM:&usg=__u_UL0DkMewZLskgjl63jV063Kkw=

please clear my doubt sir.

 

[ehild]

but if strings are placed in vertical system itself,initial momentum is mu and final momentum becomes mv+mv-mv(downwards is taken positive and the other block moves upwards).from conservation of momentum,
mu=mv
u=v.
from impulse momentum theorem and the conservation of momentum of theirs,answer is v=u/3.but according to my approach it is v=u.

but i felt conservation of linear momentum cannot be applied, as there is a net external force of 2T applied on string(which is a part of our system)due to pulley.

You are right, the linear momentum is not conserved as there is some impulsive force acting on the system.
You have to apply conservation of the angular momentum as I suggested in Post #17.

 

[ajaysabarish]

You are right, the linear momentum is not conserved as there is some impulsive force acting on the system.
You have to apply conservation of the angular momentum as I suggested in Post #17.

isn't that impulsive force caused due to pulley on the string?

but in this video,it says that momentum is conserved but they rotated the axes as shown.


note:the question is completely different but the approach is same.i have attached this video just to show how the blocks are arranged in horizontal line.

once the masses are arranged in horizontal line,they move in the same direction and final momentum becomes 3mv(in our problem).
so from conservation of momentum
mu=3mv
therefore
u=3v which is the right answer.
i couldn't understand how it is done.
please explain that sir.

 

[ehild]

isn't that impulsive force caused due to pulley on the string?

The pulley acts with both horizontal and vertical forces on the string. The vertical one can be impulsive, the horizontal one is not.

but in this video,it says that momentum is conserved but they rotated the axes as shown.

I do not see any axis rotated. What do you mean?

i have attached this video just to show how the blocks are arranged in horizontal line.

once the masses are arranged in horizontal line,they move in the same direction and final momentum becomes 3mv(in our problem).
so from conservation of momentum
mu=3mv
therefore
u=3v which is the right answer.
i couldn't understand how it is done.
please explain that sir.

The masses are not arranged in horizontal line and do not move in the same direction. Do what I said. Apply conservation of angular momentum with respect to the axes of the pulley. Do you know what angular momentum is?

 

[ajaysabarish]

The pulley acts with both horizontal and vertical forces on the string. The vertical one can be impulsive, the horizontal one is not.

why should horizontal forces act on pulley sir?only the vertical clamp force and tension force due to string act.

I do not see any axis rotated. What do you mean?

sorry sir,i just meant that blocks are arranged in horizontal line.

The masses are not arranged in horizontal line and do not move in the same direction. Do what I said. Apply conservation of angular momentum with respect to the axes of the pulley. Do you know what angular momentum is?

sir,i am not taught rotational mechanics yet,so i don't know what is angular momentum.but this problem can be solved using linear momentum concept itself sir.
sir,i very well know i am wrong but i want to know why,so please point out the mistake in the approach of applying conservation of linear momentum.

 

[ehild]

why should horizontal forces act on pulley sir?only the vertical clamp force and tension force due to string act.sorry sir,i just meant that blocks are arranged in horizontal line.
sir,i am not taught rotational mechanics yet,so i don't know what is angular momentum.but this problem can be solved using linear momentum concept itself sir.
sir,i very well know i am wrong but i want to know why,so please point out the mistake in the approach of applying conservation of linear momentum.

The string acts on the pulley and the pulley acts on the string. There is static force of friction between the pulley and the string, this results in a horizontal force. The string also presses the pulley and the pulley presses the string in vertical direction.
The blocks hang from the pulley why do you keep on saying that they are arrenged in horizontal line?
The linear momentum is not conserved.
Using angular momentum would lead to the correct solution, v=u/3.

There is an other approach people use for pulley systems. The masses move along the string. It is a one-dimensional motion, there is only forward and reverse direction. So one mass moves downward, the other one moves with the same speed upward, but both move forward along the string. You can define linear momentum along the string, then the single momenta add up as all masses move forward, so this forward momentum is 3mv. This strange momentum along the string is conserved.

 

[conscience]

respected sir,
anything can be considered as a system even a single block can be considered as a system.but we have to check whether conservation of momentum can be applied or not.so i have taken ball+blocks+string as our system.

is linear momentum conserved sir?

this is my question and my doubt.

2 blocks each of mass m are hanging as shown in the figure(https://www.google.co.in/search?q=atwood+machine&biw=1600&bih=775&tbm=isch&imgil=J-26XXHLI8jOsM%253A%253BmhX7Xm1xxTHICM%253Bhttp%25253A%25252F%25252Fphysics.stackexchange.com%25252Fquestions%25252F118917%25252FNewtons-third-law-and-atwood-machines-confusion-about-tension&source=iu&pf=m&fir=J-26XXHLI8jOsM%253A%252CmhX7Xm1xxTHICM%252C_&usg=__u_UL0DkMewZLskgjl63jV063Kkw%3D&ved=0CDIQyjdqFQoTCM_g_ZfRiskCFcSalAodIisASQ&ei=RWREVo-qBsS10gSi1oDIBA#imgrc=J-26XXHLI8jOsM%3A&usg=__u_UL0DkMewZLskgjl63jV063Kkw%3D

another ball of mass m is dropped from a height h,the ball hits one of the block and sticks to it,find the velocity of the other block after collision.

it gives 2 approaches,one is impulse momentum theorem which i understood.the second approach solves the problem using conservation of momentum,but it says momentum is conserved in the system taken(ball+blocks+string),but the string is made horizontal and the initial momentum is mu(if ball strikes the block with speed u) and final momentum is 3mv(if final speed of blocks is v).from conservation of momentum
mu=3mv
v=u/3.
but if strings are placed in vertical system itself,initial momentum is mu and final momentum becomes mv+mv-mv(downwards is taken positive and the other block moves upwards).from conservation of momentum,
mu=mv
u=v.
from impulse momentum theorem and the conservation of momentum of theirs,answer is v=u/3.but according to my approach it is v=u.

but i felt conservation of linear momentum cannot be applied, as there is a net external force of 2T applied on string(which is a part of our system)due to pulley.

please clear my doubt sir.

See,if you consider (ball+blocks+strings) as your system ,then momentum in vertical direction is not conserved .There should be no confusion regarding this .

This is how you would approach the problem .Considering downward positive ,

mu -2∫Tdt = 2mv .

∫Tdt = mv (This result is obtained by applying impulse momentum theorem separately to block which is not hit by the ball)

v = u/3 , which is exactly what you obtained with the first approach .

But this is analogous to the situation where two blocks lie on a horizontal plane (horizontal,frictionless table) connected by a taut string .An inelastic collision causes all the blocks to move together . I think the book has this setup in mind where horizontal momentum is conserved .

But it is definitely wrong to say that the momentum is conserved in vertical direction.

 

[ajaysabarish]

See,if you consider (ball+blocks+strings) as your system ,then momentum in vertical direction is not conserved .There should be no confusion regarding this .

This is how you would approach the problem .Considering downward positive ,

mu -2∫Tdt = 2mv .

∫Tdt = mv (This result is obtained by applying impulse momentum theorem separately to block which is not hit by the ball)

v = u/3 , which is what you obtained in the first approach .

But this is analogous to the situation where two blocks lie on a horizontal plane (horizontal,frictionless table) connected by a taut string .An inelastic collision causes all the blocks to move together . I think the book has this setup in mind where horizontal momentum is conserved .

But it is definitely wrong to say that the momentum is conserved in vertical direction.

thank you very for replying sir,

See,if you consider (ball+blocks+strings) as your system ,then momentum in vertical direction is not conserved .There should be no confusion regarding this .

This is how you would approach the problem .Considering downward positive ,

mu -2∫Tdt = 2mv .

∫Tdt = mv (This result is obtained by applying impulse momentum theorem separately to block which is not hit by the ball)

v = u/3 , which is exactly what you obtained with the first approach .

But this is analogous to the situation where two blocks lie on a horizontal plane (horizontal,frictionless table) connected by a taut string .An inelastic collision causes all the blocks to move together . I think the book has this setup in mind where horizontal momentum is conserved .

But it is definitely wrong to say that the momentum is conserved in vertical direction.

i almost understood it sir,but please watch this video also
so that i can be sure of the answer

 

[haruspex]

@ajaysabarish , since the pulley is ideal, the tensions in the strings are always the same on each side.
For two masses directly connected by a horizontal string, the tension acts in opposite directions on the two masses. Over some time period, therefore, the string exerts equal and opposite impulses on the two masses, so there is not net change in momentum.
With the string passing over the pulley, the two tensions act in the same direction. While the masses are hanging at rest, this is balanced by gravity, but during a sudden impulse gravity becomes insignificant, and there is a net (upward) change in momentum of the masses.
You can treat the pulley as effectively reflecting the change in momentum of the mass that is not struck from above. So if the final speed of the masses is v we have mu=mv+mv+(-m(-v)) =3mv.

 

[ajaysabarish]

@ajaysabarish , since the pulley is ideal, the tensions in the strings are always the same on each side.
For two masses directly connected by a horizontal string, the tension acts in opposite directions on the two masses. Over some time period, therefore, the string exerts equal and opposite impulses on the two masses, so there is not net change in momentum.
With the string passing over the pulley, the two tensions act in the same direction. While the masses are hanging at rest, this is balanced by gravity, but during a sudden impulse gravity becomes insignificant, and there is a net (upward) change in momentum of the masses.
You can treat the pulley as effectively reflecting the change in momentum of the mass that is not struck from above. So if the final speed of the masses is v we have mu=mv+mv+(-m(-v)) =3mv.

thank you very much sir for replying,i assume you have seen the video.i think i got it now,when the strings were vertical,the external tension impulse is in same direction(upwards),since there is a net external impulse, momentum isn't conserved,but when the strings are visualized to be horizontal,the external impulse by pulley acts on opposite sides and hence cancel out,since the net external impulse is zero,momentum is conserved.unlike the case where the strings were vertical,blocks don't move in opposite direction,rather move in same horizontal direction and hence momentum gets added up unlike the first case were strings were vertical.but is this a correct approach(considering the system in an horizontal line)?can we always consider pulley systems in a horizontal line and solve the problem?

 

[haruspex]

can we always consider pulley systems in a horizontal line and solve the problem?

I would hesitate to recommend that.

I would prefer to think in terms of the actual impulses in much the same way as one normally deals with tensions in ropes over pulleys. With an ideal pulley, the tension is the same both sides (whether the pulley is stationary or rotating, or even accelerating). In the same way, with a sudden impulse, the same upward impulse is generated each side.
Applying that view to this problem, the solo mass goes from rest to moving upwards at speed v, so the upward impulse is mv. The same upward impulse must apply to the pair of masses. This changes their total of mu down into a total of 2mv down, so the upward impulse = mu-2mv. Equating the two, mv = mu-2mv.

This way of thinking can be extended to deal with non-ideal pulleys, i.e. pulleys with nonzero moment of inertia and/or axial friction.

 

[ajaysabarish]

I would hesitate to recommend that.

I would prefer to think in terms of the actual impulses in much the same way as one normally deals with tensions in ropes over pulleys. With an ideal pulley, the tension is the same both sides (whether the pulley is stationary or rotating, or even accelerating). In the same way, with a sudden impulse, the same upward impulse is generated each side.
Applying that view to this problem, the solo mass goes from rest to moving upwards at speed v, so the upward impulse is mv. The same upward impulse must apply to the pair of masses. This changes their total of mu down into a total of 2mv down, so the upward impulse = mu-2mv. Equating the two, mv = mu-2mv.

This way of thinking can be extended to deal with non-ideal pulleys, i.e. pulleys with nonzero moment of inertia and/or axial friction.

thank you very much sir for replying,i thought the same but i wanted to know why this approach works.
but is this explanation for the approach correct sir?

when the strings were vertical,the external tension impulse is in same direction(upwards),since there is a net external impulse, momentum isn't conserved,but when the strings are visualized to be horizontal,the external impulse by pulley acts on opposite sides and hence cancel out,since the net external impulse is zero,momentum is conserved.unlike the case where the strings were vertical,blocks don't move in opposite direction,rather move in same horizontal direction and hence momentum gets added up unlike the first case were strings were vertical

 

[haruspex]

thank you very much sir for replying,i thought the same but i wanted to know why this approach works.
but is this explanation for the approach correct sir?

when the strings were vertical,the external tension impulse is in same direction(upwards),since there is a net external impulse, momentum isn't conserved,but when the strings are visualized to be horizontal,the external impulse by pulley acts on opposite sides and hence cancel out,since the net external impulse is zero,momentum is conserved.unlike the case where the strings were vertical,blocks don't move in opposite direction,rather move in same horizontal direction and hence momentum gets added up unlike the first case were strings were vertical

Yes.

 

[ajaysabarish]

Yes.

sir i hope you have watched the video because i don't think i conveyed the method correctly through text.in the video it is shown how system is arranged in horizontal line and how external impulse gets cancelled.

 

[ehild]

@ajaysabarish,
Haruspex gave you a very good explanation, but using quantities which are conserved provides an easy solution for problems. In case there is a fixed axis of rotation, momentum is not conserved, but the angular momentum does if there is no external torque. If a point mass m performs uniform circular motion with speed v and the radius is r, its velocity changes all the time, so is its momentum, but its angular momentum L=mrv is constant.
I hope you will learn about angular momentum soon.
The angular momentum with respect to a point is m rXv, where r is the position vector and v is the velocity. X means cross-product, It is easy to calculate as multiplying the speed with the distance of the lline of velocity from the axes of rotation. The sign is positive if the motion corresponds to an anti-clockwise rotation about the axis and negative when it is clock-wise. The same way as it is with the torque, I hope you are familiar with it, or you will learn it soon.
The angular momentum of a system is conserved if there are no external torques or they can be ignored during the short time of collision. It is the case here, as the only torques present are because of gravity.
If the pulley has radius r, the string is at distance r from the rotation axis. If the falling body falls along the string with speed v, its angular momentum is Li = mrv. As the other masses are in rest, Li is the total initial angular momentum of the system. Li is positive as it corresponds anti-clockwise rotation of the pulley.
After the collision, all masses move with speed u, the sticked together masses downward with angular momentum 2mur, and the other mass upward on the other side. It also corresponds to anticlockwise rotation of the pulley, so that angular momentum is also positive, .So the final angular momentum is Lf=2mru+mru=3mru.
Lf=Li so 3mru=mrv. Divide the equation with r. You see, no need to imagine that the masses are arranged horizontally and no need to work with the impulse.

 

[ajaysabarish]

@ajaysabarish,
Haruspex gave you a very good explanation, but using quantities which are conserved provides an easy solution for problems. In case there is a fixed axis of rotation, momentum is not conserved, but the angular momentum does if there is no external torque. If a point mass m performs uniform circular motion with speed v and the radius is r, its velocity changes all the time, so is its momentum, but its angular momentum L=mrv is constant.
I hope you will learn about angular momentum soon.
The angular momentum with respect to a point is m rXv, where r is the position vector and v is the velocity. X means cross-product, It is easy to calculate as multiplying the speed with the distance of the lline of velocity from the axes of rotation. The sign is positive if the motion corresponds to an anti-clockwise rotation about the axis and negative when it is clock-wise. The same way as it is with the torque, I hope you are familiar with it, or you will learn it soon.
The angular momentum of a system is conserved if there are no external torques or they can be ignored during the short time of collision. It is the case here, as the only torques present are because of gravity.
If the pulley has radius r, the string is at distance r from the rotation axis. If the falling body falls along the string with speed v, its angular momentum is Li = mrv. As the other masses are in rest, Li is the total initial angular momentum of the system. Li is positive as it corresponds anti-clockwise rotation of the pulley.
After the collision, all masses move with speed u, the sticked together masses downward with angular momentum 2mur, and the other mass upward on the other side. It also corresponds to anticlockwise rotation of the pulley, so that angular momentum is also positive, .So the final angular momentum is Lf=2mru+mru=3mru.
Lf=Li so 3mru=mrv. Divide the equation with r. You see, no need to imagine that the masses are arranged horizontally and no need to work with the impulse.

thank you very much sir for replying,yes i don't know anything about angular momentum and torque but i will learn it soon and after learning about those concepts i will definitely revisit this thread and see your method too.thank you for providing another method sir.