A baseball batter hits an incoming 45.0-m/s fastball.The ball leaves the bat at 56.0 m/s after a ball-on-bat contact time of 0.040 s.What is the force exerted on the 0.25-kg baseball?
A)631 N
B)68.8 N
C)350 N
D)16.2 N
E)281 N

Question

Quizplus · Accepted Answer

The force exerted on the baseball can be calculated using Newton's second law, $F = ma$, where $m$ is the mass of the baseball and $a$ is the acceleration. The acceleration can be found using the change in velocity ($\Delta v$) over the time ($\Delta t$) the force was applied. The change in velocity is the final velocity minus the initial velocity, which is $56.0 \, \text{m/s} - (-45.0 \, \text{m/s}) = 101.0 \, \text{m/s}$ (since the initial velocity is in the opposite direction, it's considered negative). The time is $0.040 \, \text{s}$. Thus, $a = \frac{\Delta v}{\Delta t} = \frac{101.0 \, \text{m/s}}{0.040 \, \text{s}} = 2525 \, \text{m/s}^2$. The force is then $F = ma = 0.25 \, \text{kg} \times 2525 \, \text{m/s}^2 = 631.25 \, \text{N}$, which rounds to $631 \, \text{N}$.

A Baseball Batter Hits an Incoming 45