Hey there! Sign in to join this conversationNew here? Join for free
    • Thread Starter
    Offline

    2
    ReputationRep:
    Particles A, B and C of masses 4m, 3m and m respectively, lie at rest in a straight horizontal plane with B between A and C. Particles A and B are projected towards each other with speeds u ms^{-1} and v ms^{-1} respectively, and collide directly.

    As a result of the collision, A is brought to rest and B is rebounded with speed kv ms^{-1}. The coefficient of restitution between A and B is \frac{3}{4}.

    a) Show that u = 3v. (6)

    b) Find the value of k. (2)

    Immediately after the collision between A and B, particle C is projected with speed 2v ms^{-1} towards B so that B and C collide directly.

    c) Show that there is no further collision between A and B. (4)
    My workings so far (Diagrams are attatched):

    a)
    Conservation of momentum:

    4mu - 3mv = 3mkv

    Newton's Law of Restitution:

    e = \frac{separation speed}{approach speed}
    \frac{3}{4} = \frac{3mkv}{4mu + 3mv}

    Now I am unsure as to what I can do next??

    b)
     u = 3v

    12mv -3mv = 3mkv

    k = 3

    c)
    Conservation of momentum:

    9mv - 2mv = 3mv^{B} + mv^{C}

    Again I am unsure as to what I do next?? It's obvious that B won't collide with A again as B will keep moving in the same direction after hitting C but I can't see how to show that??
    Attached Images
      
    Offline

    1
    ReputationRep:
    (Original post by JamesyB)
    Newton's Law of Restitution:

    e = \frac{separation speed}{approach speed}
    \frac{3}{4} = \frac{3mkv}{4mu + 3mv}

    Now I am unsure as to what I can do next??
    I think you are misunderstanding the law of restitution. It relates to speed (as you said above) not momentum (as you used below it).

    You can also divide your momentum equations through by m as it's the same value for all of them. This will leave you with some simultaneous equations which you should be able to solve.

    (Original post by JamesyB)
    c)
    Conservation of momentum:

    9mv - 2mv = 3mv^{B} + mv^{C}

    Again I am unsure as to what I do next?? It's obvious that B won't collide with A again as B will keep moving in the same direction after hitting C but I can't see how to show that??
    Don't know if it's the most elegant solution, but if you use restitution as in part a (using e), and solve simultaneously again, you should get a speed of B after the collision (in terms of v and e). The condition this velocity needs to satisfy is (taking away from a as positive) Velocity_B \geq 0. If you find it in terms of v and e, and use e as 0 \leq e \leq 1, it should be clear this is positive (you'll see what I mean if you get it)

    Sorry if that's not too clear. If you need a more explanation on part of it, ask.
    • Thread Starter
    Offline

    2
    ReputationRep:
    (Original post by gcseeeman)
    I think you are misunderstanding the law of restitution. It relates to speed (as you said above) not momentum (as you used below it).

    You can also divide your momentum equations through by m as it's the same value for all of them. This will leave you with some simultaneous equations which you should be able to solve.



    Don't know if it's the most elegant solution, but if you use restitution as in part a (using e), and solve simultaneously again, you should get a speed of B after the collision (in terms of v and e). the condition this speed needs to satisfy is (taking away from a as positive) Velocity_B \geq 0. If you find it in terms of v and e, and use e as 0 \leq e \leq 1, it should be clear this is positive (you'll see what I mean if you get it)

    Sorry if that's not too clear. If you need a more explanation on part of it, ask.
    Ah cheers, mechanics is really not my strong point haha.

    Got both parts now
 
 
 
  • See more of what you like on The Student Room

    You can personalise what you see on TSR. Tell us a little about yourself to get started.

  • Poll
    Has a teacher ever helped you cheat?
    Useful resources

    Make your revision easier

    Maths

    Maths Forum posting guidelines

    Not sure where to post? Read the updated guidelines here

    Equations

    How to use LaTex

    Writing equations the easy way

    Student revising

    Study habits of A* students

    Top tips from students who have already aced their exams

    Study Planner

    Create your own Study Planner

    Never miss a deadline again

    Polling station sign

    Thinking about a maths degree?

    Chat with other maths applicants

    Can you help? Study help unanswered threads

    Groups associated with this forum:

    View associated groups
  • See more of what you like on The Student Room

    You can personalise what you see on TSR. Tell us a little about yourself to get started.

  • The Student Room, Get Revising and Marked by Teachers are trading names of The Student Room Group Ltd.

    Register Number: 04666380 (England and Wales), VAT No. 806 8067 22 Registered Office: International House, Queens Road, Brighton, BN1 3XE

    Write a reply...
    Reply
    Hide
    Reputation gems: You get these gems as you gain rep from other members for making good contributions and giving helpful advice.