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Math Help - Newton's Law Of Restitution

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    Newton's Law Of Restitution

    I've done part (a) but cannot understand the method of how part (b) is done. I've written down the answer I obtained (which is correct) to part (a). Can someone help? Thanks in advance.

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    Q:

    A small smooth ball A of mass m is moving on a horizontal table with speed u when it collides directly with another small smooth ball B of mass 3m which is at rest on the table. The balls have the same radius and the coefficient of restitution between the balls is e. The direction of motion of A is reversed as a result of the collision.

    (a) Find, in terms of e and u. the speeds of A and B immediately after the collision.


    In the subsequent motion B strikes a vertical wall, which is perpendicular to the direction of motion of B, and rebounds. The coefficient of restitution between B and the wall is \frac{3}{4}.

    Given that there is a second collision between A and B,

    (b) find the range of values of e for which the motion described is possible.

    ____________________
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    A:


    (a)
    v_1 = \frac{1}{4} u (3e - 1)
    v_2 = \frac{1}{4}u(e + 1)
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    Quote Originally Posted by Air View Post
    I've done part (a) but cannot understand the method of how part (b) is done. I've written down the answer I obtained (which is correct) to part (a). Can someone help? Thanks in advance.

    ____________________
    ____________________

    Q:

    A small smooth ball A of mass m is moving on a horizontal table with speed u when it collides directly with another small smooth ball B of mass 3m which is at rest on the table. The balls have the same radius and the coefficient of restitution between the balls is e. The direction of motion of A is reversed as a result of the collision.

    (a) Find, in terms of e and u. the speeds of A and B immediately after the collision.


    In the subsequent motion B strikes a vertical wall, which is perpendicular to the direction of motion of B, and rebounds. The coefficient of restitution between B and the wall is \frac{3}{4}.

    Given that there is a second collision between A and B,

    (b) find the range of values of e for which the motion described is possible.

    ____________________
    ____________________


    A:


    (a)
    v_1 = \frac{1}{4} u (3e - 1)
    v_2 = \frac{1}{4}u(e + 1)
    So let the +x direction be the direction of the motion of ball A before the initial collision. Conservation of momentum says:
    mu = -mv_A + 3mv_B
    (I'm taking the first term on the right to be negative since the problem stipulates that A is rebounding.)

    We also have
    e = \frac{v_A + v_B}{u}

    You can solve this for, say v_B and get that
    v_B = eu - v_A

    Plug that into your momentum equation and you can find v_A, and then use either equation to find v_B. Given this method, I agree with your answer to part a).

    For part b), object B strikes the wall and rebounds in the negative direction with a speed v_B^{\prime}. Since we need ball B to strike ball A again, v_B^{\prime} > v_A is your criterion.

    So what's v_B^{\prime}? Well the coefficient of restitution for the collision with the wall is 3/4. So
    \frac{3}{4} = \frac{v_B^{\prime}}{v_B}

    v_B^{\prime} = \frac{3}{4} \cdot v_B = \frac{3}{4} \cdot \frac{1}{4}u(e + 1) = \frac{3}{16} \cdot u(e + 1)

    So solve
    \frac{3}{16} \cdot u(e + 1) > \frac{1}{4} u (3e - 1)
    for e.

    -Dan
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