Question

A 0.145-kg baseball is traveling to the left toward home plate with a véiocity of 40 m/s. After

getting hit by a bat, it travels 50 m/s to the right. What is the baseball's change in momentum?
(Hint: remember that the change in momentum is = Pfinal - Pinitial-)

Answer 1

`\Delta p=13.05\ kg.m/s`

Step-by-step explanation:

Change in momentum

The momentum of a body of mass m and velocity v is given by:

`\vec p=m\vec v`

Both the momentum and velocity are vectors. The change in momentum is:

`\Delta \vec p=\vec p_2-\vec p_1`

Where
`\vec p_2, \vec p_1` are the final and initial momentums.

If the object is restricted to move in one dimension, then the direction of the motion is considered with the signs of the velocity and/or the momentum.

Let's consider the positive velocity in the right direction and the negative velocity in the left direction.

The baseball with mas m=0.145 kg is initially traveling to the left with a velocity:

`v_1=-40 \ m/s`

The initial momentum is:

`p_1=mv_1=0.145*(-40)=-5.8\ kg.m/s`

After getting hit by the bat, it travels at 50 m/s to the right, thus:

`v_2=50 \ m/s`

The initial momentum is:

`p_2=mv_2=0.145*(50)=7.25\ kg.m/s`

The change in momentum is

`\Delta p=7.25\ kg.m/s-(-5.8\ kg.m/s)`

`\mathbf{\Delta p=13.05\ kg.m/s}`