Question

A 10.0kg ball traveling at 13.0m/s is crashed into from behind by a 11.0kg ball traveling at 22.0m/s in the same direction. What is the velocity of the first ball if the second slows down to 19.2m/s after the collision?

Answer 1

In order to calculate the velocity of the first ball after the collision, we can use the equation for the conservation of momentum:

$m_1v_{1i}+m_2v_{2i}=m_1v_{1f}+m_2v_{2f}$

Where m1 and m2 are the masses, v1i and v2i are the initial velocities and v1f and v2f are the final velocities.

So, using the given values, we have:

$\begin{gathered} 10\cdot13+11\cdot22=10\cdot v_{1f}+11\cdot19.2 \\ 130+242=10v_{1f}+211.2 \\ 372-211.2=10v_{1f} \\ 10v_{1f}=160.8 \\ v_{1f}=16.08\text{ m/s} \end{gathered}$

Therefore the velocity of the first ball after the collision is 16.08 m/s.