Question

What is the average speed of the rail + bowling balls system sometime after the collision?
A. $\dfrac{{\left( {M + Nm} \right){v_0}}}{M}$
B. $\dfrac{{M{v_0}}}{{\left( {Nm + M} \right)}}$
C. $\dfrac{{Nm{v_0}}}{M}$
D. The speed cannot be determined because there is not enough information.

Answer 1

In order to answer this question first we will find the final velocity of the railcar + bowling ball in the horizontal direction. After that the balls, on the other hand, move vertically, where momentum is not conserved, and their final velocity is determined by the value of time $t$ , after which the system's velocity is considered.

Complete step by step answer:
In this question, because there is no external force in this direction, the momentum component parallel to the track will be conserved. Therefore assume that the final horizontal speed of the railcar + bowling balls is $V$ . Then, using conservation of linear momentum, one can write,
$M \times {v_0} = \left( {N \times m + M} \right) \times V \\\ \Rightarrow V = \dfrac{{M \times {v_0}}}{{\left( {N \times m + M} \right)}}$
Here, $M$ is the mass of the rail car and ${v_0}$ is the initial speed, $N$ is the number of bowling balls, each of mass $m$.

The balls, on the other hand, move vertically, where momentum is not conserved, and their final velocity is determined by the value of time t, after which the system's velocity is considered.As a result, the average speed of the railcar + bowling balls system cannot be calculated without any assumptions, but if the vertical component of momentum of the bouncing balls (some time after collision) is assumed to be negligible in comparison to the horizontal momentum of the system, the average speed of the system can be equated to $\dfrac{{M \times {v_0}}}{{\left( {N \times m + M} \right)}}$.

So, the correct option is B.

Note: Only the component of momentum parallel to the track is preserved. Internal forces are greater than external gravity forces when the balls crash with the train bed, and because the action-reaction pair cancels out the influence of internal forces, the rate of change of momentum for the colliding balls is zero. In the vertical direction, this breaks down the conservation of momentum.