If we push one cart (Cart A) into a second stationary cart (Cart B), v_Ai is nonzero and v_Bi=0, respectively. Assume no outside forces will

Question

If we push one cart (Cart A) into a second stationary cart (Cart B), v_Ai is nonzero and v_Bi=0, respectively. Assume no outside forces will be involved and the collision is elastic (i.e. both momentum and kinetic energy are conserved). Derive the general equation for the velocity of Cart A after the collision v_Af as well as the velocity of Cart B after the collision v_Bf. We want these to be in terms of known quantities: initial velocity v_Ai and the mass of the carts m_A and m_B. Refer to the “Useful Theory” section below to get you started. If you get stuck, work with your lab partners and refer to your text to find the solution before

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Ben Gia 4 years 2021-08-12T06:43:29+00:00 1 Answers 12 views 0

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  1. Nem
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    2021-08-12T06:44:46+00:00
    Reply

    Answer:

    v_A=(\frac{m_A-m_B}{m_A+m_B})u_A

    v_B=(\frac{2m_A}{m_A+m_B})u_A

    Explanation:

    In an elastic collision, the total momentum of the system and the total kinetic energy of the system are conserved.

    So:

    – Conservation of momentum:

    m_A u_A+m_B u_B = m_A v_A + m_B v_B (1)

    where

    m_A, m_B are the masses of cart A and B

    u_A,u_B are the initial velocities of cart A and B

    v_A,v_B are the final velocities of cart A and B

    – Conservation of kinetic energy:

    \frac{1}{2}m_A u_A^2+\frac{1}{2}m_B u_B^2= \frac{1}{2}m_A v_A^2 +\frac{1}{2} m_B v_B^2 (2)

    From (1) we can write:

    m_A (u_A-v_A)=m_B(v_B-u_B)\\\frac{m_A (u_A-v_A)}{m_B(v_B-u_B)}=1 (3)

    From (2) we get:

    m_A (u_A^2-v_A^2)=m_B(v_B^2-u_B^2)\\m_A(u_A-v_A)(u_A+v_A)=m_B(v_B-u_B)(v_B+u_B)\\\frac{m_A(u_A-v_A)}{m_B(v_B-u_B)}(u_A+v_A)=(v_B+u_B) (4)

    Substituting (3) into (4),

    1\cdot (u_A+v_A)=(v_B+u_B)\\v_B=u_A+v_A-u_B (5)

    And substituting (5) into (1),

    m_A(u_A-v_A)=m_B (u_A+v_A-2u_B)

    And now we can solve to find the final expressions:

    v_A=(\frac{m_A-m_B}{m_A+m_B})u_A+(\frac{2m_B}{m_A+m_B})u_B

    v_B=(\frac{2m_A}{m_A+m_B})u_A+(\frac{m_B-m_A}{m_A+m_B})u_B

    But in this problem

    u_B=0

    So the equations can be rewritten as

    v_A=(\frac{m_A-m_B}{m_A+m_B})u_A

    v_B=(\frac{2m_A}{m_A+m_B})u_A

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