Question

1. a baseball of mass 0.145 kg is thrown at 45.0 m/s and is batted away at 50.0 m/s by the batter with a 1.00 kg bat after being in contact with the bat for 1.00 ms. what is the acceleration of the batters bat?

Explanation:

Step1: Convert time unit

Convert \(1.00\ ms\) to \(s\). Since \(1\ ms = 10^{- 3}\ s\), then \(t=1.00\times10^{-3}\ s\).

Step2: Use the formula \(a=\frac{\Delta v}{t}\)

The initial velocity of the bat \(v_{0} = 0\ m/s\) (assuming the bat is at rest before hitting, this is a common assumption in such problems as the problem focuses on the change due to the hit - like situation). The final velocity of the bat is not relevant for calculating its acceleration from rest in the context of the contact time (if we assume the bat starts from rest and we are calculating the acceleration during the contact time). But if we consider the force - impulse relation, we can also use the fact that the force on the ball \(F=\frac{m_{ball}(v_{final,ball}-v_{initial,ball})}{t}\), and by Newton's third law \(F = m_{bat}a\).

First, calculate the force on the ball. \(m_{ball}=0.145\ kg\), \(v_{initial,ball}=- 45.0\ m/s\) (assuming the initial direction of the ball is opposite to the final direction), \(v_{final,ball}=50.0\ m/s\), \(t = 1.00\times10^{-3}\ s\).

\(F=\frac{m_{ball}(v_{final,ball}-v_{initial,ball})}{t}=\frac{0.145\times(50.0 + 45.0)}{1.00\times10^{-3}}\)

\(F=\frac{0.145\times95.0}{1.00\times10^{-3}}=\frac{13.775}{1.00\times10^{-3}}=13775\ N\)

Since \(F = m_{bat}a\) and \(m_{bat}=1.00\ kg\), then \(a=\frac{F}{m_{bat}}\)

\(a=\frac{13775}{1.00}=13775\ m/s^{2}\)

Answer:

\(13775\ m/s^{2}\)