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.