Question

I need help with part a of question number four

Answer 1

Given that mass of the baseball is, M = 0.142 kg and the mass of the bullet is m = 2.62 g = 2.62 x 10^(-3) kg and the speed of the bullet is, v = 1.5 x 10^3 m/s

Also, the momentum of both the baseball and bullet are equal, then the speed of the baseball, v' will be

`\begin{gathered} Mv^(\prime)=mv \\ v^(\prime)=(mv)/(M) \\ v^(\prime)=\text{ 27.67 m/s} \end{gathered}`

The speed of the baseball is 27.67 m/s.