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Can conservation of momentum be applied here?

  1. Nov 12, 2015 #1
    • Member warned to use the formatting template for homework posts.
    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.
     
    Last edited: Nov 12, 2015
  2. jcsd
  3. Nov 12, 2015 #2

    ehild

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    Without the pull,y it is not an Atwood machine.
     
  4. Nov 12, 2015 #3
    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.
     
  5. Nov 12, 2015 #4
    Is the ball moving vertically or horizontally before colliding with the block ?
     
  6. Nov 12, 2015 #5
    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.
     
  7. Nov 12, 2015 #6
    Do you think there are any vertical forces acting on the system (blocks+ball+string) during the collision ?
     
  8. Nov 12, 2015 #7
    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.
     
  9. Nov 12, 2015 #8
    Do you think there would be any tension in the strings if there was no gravity ?
     
  10. Nov 12, 2015 #9
    yes sir i forgot that,i will edit the question and i am sorry for wasting your time.
     
  11. Nov 12, 2015 #10
    now i have changed the question sir,please clear my doubt.
     
  12. Nov 12, 2015 #11
    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 ?
     
  13. Nov 12, 2015 #12

    ehild

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    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?
     
  14. Nov 12, 2015 #13
    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) .
     
  15. Nov 12, 2015 #14
    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.
     
  16. Nov 12, 2015 #15
    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.
     
  17. Nov 12, 2015 #16
    yes sir,pulley applies a force on our system but pulley is not considered to be a part of our system(ball+blocks+string)
     
  18. Nov 12, 2015 #17

    ehild

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    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 Fi and it is not true, that Fidt 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.
     
  19. Nov 12, 2015 #18
    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?
     
  20. Nov 12, 2015 #19
    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.
     
  21. Nov 12, 2015 #20

    ehild

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    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.
     
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