Question

A pitching machine throws a 0.148kg baseball to the left at 26m/s. A batter then hits the baseball directly back at 39m/s. What is the change in momentum of this baseball

Answer 1

Let us assume that the initial direction of the ball, i.e., from the pitching machine to the batter as the negative direction, and the final direction of the ball is the positive direction.

Given,

The mass of the baseball, m=0.148 kg

The initial velocity of the baseball, u=-26 m/s

The final velocity of the baseball, v=39 m/s

The momentum of the ball before the batter hits it is given by,

$p_1=mu$

On substituting the known values,

$\begin{gathered} p_1=0.148\times-26 \\ =-3.848\text{ kg}\cdot\text{m/s} \end{gathered}$

The momentum of the ball after the batter hits it is given by,

$p_2=mv$

On substituting the known values,

$\begin{gathered} p_2=0.148\times39 \\ =5.772\text{ kg}\cdot\text{m/s} \end{gathered}$

The change in the momentum is given by,

$\Delta p=p_2-p_1$

On substituting the known values,

$\begin{gathered} \Delta p=5.772-(-3.848) \\ =9.62\text{ kg}\cdot\text{m/s} \end{gathered}$

Thus the change in the momentum of baseball is 9.62 kg·m/s