Question

Part C: What was the ball's change in the momentum during its impact with the floor?

Answer 1

Given,

The mass of the ball, m=280 g=0.28 kg

The velocity of the ball just before it hit the ground, u=-2.5 m/s

The velocity of the ball while rebounding, v=2.0 m/s

The momentum of an object is given by the product of the mass of the object and the velocity of the object.

Part A:

The momentum of the ball before hitting the ground,

`p_1=mu`

On substituting the known values,

`\begin{gathered} p_1=0.28*-2.5 \\ =-0.70\text{ kg}\cdot(m)/(s) \end{gathered}`

Thus the momentum of the ball just before it hit the ground is -0.70 kg·m/s

Part B;

The momentum of the ball just after it rebounds is,

`p_2=mv`

On substituting the known values,

`\begin{gathered} p_2=0.28*2.0 \\ =0.56\text{ kg}\cdot\text{m/s} \end{gathered}`

Thus the momentum of the ball just after it rebounded from the floor is 0.56 kg·m/s

Part C:

The change in the momentum is given by,

`\Delta p=p_2-p_1`

On substituting the known values,

`\begin{gathered} \Delta p=0.56-(-0.7) \\ =1.26\text{ kg}\cdot\text{m/s} \end{gathered}`

Thus the change in the ball's momentum during its impact with the floor is 1.26 kg·m/s