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

The change in momentum of the baseball is 9.620 kg.m/s.

Step-by-step explanation:

The change in momentum of an object is equal to the final momentum minus the initial momentum. The momentum of an object is calculated by multiplying its mass by its velocity.

In this case, the initial momentum of the baseball is given by the product of its mass (0.148 kg) and its initial velocity (-26 m/s) since it's thrown to the left. The final momentum is given by the product of the same mass and its final velocity (39 m/s) in the opposite direction.

To calculate the change in momentum, subtract the initial momentum from the final momentum:

Change in momentum = Final momentum - Initial momentum

Change in momentum = (0.148 kg imes 39 m/s) - (0.148 kg imes -26 m/s)

Change in momentum = 5.772 kg.m/s + 3.848 kg.m/s = 9.620 kg.m/s

Answer 2

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*-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*39 \\ =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