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Math Help - Newton's Law Of Restitution

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    Newton's Law Of Restitution

    A smooth sphere P of mass 2m is moving in a straight line with speed u on a smooth horizontal table. Another smooth sphere Q of mass m is at rest on the table. The sphere P collides directly with Q. The coefficient of restitution between P and Q is \frac{1}{3}. The spheres are modelled as particles.

    (a) Show that, immediately after the collision, the speeds of P and Q are \frac{5}{9}u and \frac{8}{9}u respectively.

    After the collision, Q strikes a fixed vertical wall which is perpendicular to the direction of motion of P and Q. The coefficient of restitution between Q and the wall is e. When P and Q collide again, P is brought to rest.

    (b) Find the value of e.

    (c) Explain why there must be a third collision between P and Q.

    ____________________
    ____________________


    I've done part (a) by forming 2 simultaneous equations of conservation of momentum and Newton's law of restitution and solving to obtain the values required. For part (c), I know that collusion must occur as particle P is rebounded so it will hit the stationary particle Q.

    The only aspect I don't understand is part (b). How would I do this? Thanks in advance.
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    Quote Originally Posted by Air View Post
    A smooth sphere P of mass 2m is moving in a straight line with speed u on a smooth horizontal table. Another smooth sphere Q of mass m is at rest on the table. The sphere P collides directly with Q. The coefficient of restitution between P and Q is \frac{1}{3}. The spheres are modelled as particles.

    (a) Show that, immediately after the collision, the speeds of P and Q are \frac{5}{9}u and \frac{8}{9}u respectively.

    After the collision, Q strikes a fixed vertical wall which is perpendicular to the direction of motion of P and Q. The coefficient of restitution between Q and the wall is e. When P and Q collide again, P is brought to rest.

    (b) Find the value of e.

    (c) Explain why there must be a third collision between P and Q.

    ____________________
    ____________________


    I've done part (a) by forming 2 simultaneous equations of conservation of momentum and Newton's law of restitution and solving to obtain the values required. For part (c), I know that collusion must occur as particle P is rebounded so it will hit the stationary particle Q.

    The only aspect I don't understand is part (b). How would I do this? Thanks in advance.
    We have the second impact; Q against the wall. So calling the speed of Q before it hits the wall p_Q we have
    e = \frac{v_Q^{\prime}}{v_Q}
    where v_Q^{\prime} is the magnitude of the velocity of Q after the collision with the wall. (Obviously this velocity is now opposite to the velocity of P.)

    Thus
    v_Q^{\prime} = ev_Q = \frac{8}{9}eu

    Now we have the second collision with P. Set this up the same way as before: with a conservation of momentum equation and the coefficient of restitution is again 1/3. Combining these two equations I get
    2mv_P -mv_Q^{\prime} = \frac{1}{27}(5 + 8e)mu
    which you can solve for e.

    -Dan
    Last edited by topsquark; May 18th 2008 at 06:45 AM.
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    I redid this question and I am slightly confused. This picture that I am seeing is this:



    So, my working is:

    e = \frac{w}{\frac{8u}{9}} \implies \frac{8}{9}ue = w

    \text{COM:} \ \left( \frac{5}{9}u \right) \left( 2m \right) + \left(  \frac{8}{9}u \right)  \left( m \right)  = - \left( \frac{8}{9}ue \right)  \left( m \right)

    \frac{10}{9} + \frac{8}{9} = -\frac{8}{9}e

    e = - \frac{9}{4}.

    The answer is wrong. What have I done wrong? topsquark, I don't understand why it is negative when before it is travelling in the positive direction. The correct answer is e = \frac{25}{32}.
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    Forum Admin topsquark's Avatar
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    Quote Originally Posted by Air View Post
    I redid this question and I am slightly confused. This picture that I am seeing is this:



    So, my working is:

    e = \frac{w}{\frac{8u}{9}} \implies \frac{8}{9}ue = w

    \text{COM:} \ \left( \frac{5}{9}u \right) \left( 2m \right) + \left(  \frac{8}{9}u \right)  \left( m \right)  = - \left( \frac{8}{9}ue \right)  \left( m \right)

    \frac{10}{9} + \frac{8}{9} = -\frac{8}{9}e
    The positive direction in this problem has been defined to be the initial direction of P: to the right. So Q is moving to the right before it hits the wall: u is positive. When it rebounds from the wall Q is moving to the left: w is negative.

    -Dan
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    Quote Originally Posted by topsquark View Post
    The positive direction in this problem has been defined to be the initial direction of P: to the right. So Q is moving to the right before it hits the wall: u is positive. When it rebounds from the wall Q is moving to the left: w is negative.

    -Dan
    But when I did that with conservation of momentum, it doesn't appear to work:

    \text{COM:} \ \left( \frac{5}{9}u \right) \left( 2m \right) + \left( \frac{8}{9}u \right) \left( m \right) = - \left( \frac{8}{9}ue \right) \left( m \right)

    ^ See, I put negative \left( \frac{8}{9}ue \right).
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    PFX
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    Quote Originally Posted by Air View Post
    After the collision, Q strikes a fixed vertical wall which is perpendicular to the direction of motion of P and Q. The coefficient of restitution between Q and the wall is e. When P and Q collide again, P is brought to rest. (b) Find the value of e.
    to find e we use
    e = w/(8u/9)


    first we should find w
    therefore

    2m (5u/9)+m(-w) = 2m(0)+m(v) (1)

    where w is rebound velocity of Q after it hits the wall and v is its velocity after the second collision with P

    and another equation

    e = 1/3 (between P and Q)

    therefore
    1/3 = (v-0)/ ((5u/9)+w) (2)

    from eq. 1
    v = (10u/9) - w (3)

    substituting the value of v from eq. 3 into eq. 2 and solving simultaneously...

    this gives us
    w = 25u/36
    and
    v = 5u/12

    e = (25u/36)/(8u/9)

    which gives e = 25/32
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