Source: High school physics (Chinese)
Problem Sets:
Problem
A 0.14 kg softball is pitched horizontally at 50 m/s. It is hit by a bat and moves back at a speed of 80 m/s at an angle of 45° above the horizontal.
- Find the magnitude and direction of the impulse delivered by the bat.
- If the ball-bat contact time is 0.02 s, find the magnitude of the average force exerted by the bat.
- How many times the softball's weight is this force?
[Q1] $I = 16.9$ N s, at 28.0° above the reversed pitch direction. [Q2] $F_{avg} = 845$ N. [Q3] 616 times the weight.
Let the initial velocity be along the negative x-axis, $\vec{v}_i = (-50, 0)$ m/s. The final velocity is $\vec{v}_f = (80\cos 45°, 80\sin 45°)$ m/s. During the brief impact, the impulse due to gravity is negligible compared to the impulse from the bat. Therefore, the impulse from the bat is approximately the total change in momentum: $\vec{I}_{bat} \approx \Delta\vec{p} = m(\vec{v}_f - \vec{v}_i)$. The components of the impulse are:
$$I_x = m(v_{fx} - v_{ix}) = 0.14(80\cos 45° - (-50))$$ $$I_y = m(v_{fy} - v_{iy}) = 0.14(80\sin 45° - 0)$$The magnitude is $|I| = \sqrt{I_x^2 + I_y^2}$ and the direction is $\theta = \arctan(I_y/I_x)$. The average force is $\bar{F}_{bat} = |\vec{I}_{bat}| / \Delta t$. The ratio is $N = \bar{F}_{bat} / (mg)$.