Question

A 155-g baseball is incoming at a velocity of 25 m/s. The batter hits the ball as shown in the image. The outgoing baseball has a velocity of 20 m/s at the angle 60 .

find the magnitude of the impulse acting on the ball during the hit?

Answer 1

The magnitude of the impulse acting on the ball during the hit is approximately
`\(0.775 \, \text{Ns}\).`

To find the magnitude of the impulse acting on the ball, you can use the impulse-momentum theorem, which states:

`\[ \text{Impulse} = \Delta p \]`

where
`\( \Delta p \)` is the change in momentum.

The momentum (p) is given by the product of mass (m) and velocity (v):

`\[ p = m \cdot v \]`

The change in momentum is then:

`\[ \Delta p = p_{\text{final}} - p_{\text{initial}} \]`

For the incoming baseball:

`\[ p_{\text{initial}} = m_{\text{incoming}} \cdot v_{\text{incoming}} \]`

For the outgoing baseball:

`\[ p_{\text{final}} = m_{\text{outgoing}} \cdot v_{\text{outgoing}} \]`

The magnitude of the impulse is the absolute value of the change in momentum:

`\[ \text{Impulse} = |\Delta p| \]`

Given:

-
`\( m_{\text{incoming}} = 155 \, \text{g} \) (convert to kg: \(0.155 \, \text{kg}\)),`

-
`\( v_{\text{incoming}} = 25 \, \text{m/s} \),`

-
`\( m_{\text{outgoing}} = 155 \, \text{g} \) (convert to kg: \(0.155 \, \text{kg}\)),`

-
`\( v_{\text{outgoing}} = 20 \, \text{m/s} \).`

Calculate
`\( p_{\text{initial}} \), \( p_{\text{final}} \), and then \( \Delta p \):`

`\[ p_{\text{initial}} = 0.155 \, \text{kg} \cdot 25 \, \text{m/s} \]`

`\[ p_{\text{initial}} = 3.875 \, \text{Ns} \]`

`\[ p_{\text{final}} = 0.155 \, \text{kg} \cdot 20 \, \text{m/s} \]`

`\[ p_{\text{final}} = 3.1 \, \text{Ns} \]`

`\[ \Delta p = |3.1 \, \text{Ns} - 3.875 \, \text{Ns}| \]`

`\[ \Delta p \approx 0.775 \, \text{Ns} \]`

Therefore, the magnitude of the impulse acting on the ball during the hit is approximately
`\(0.775 \, \text{Ns}\).`