Question

An official major league baseball has a mass of 0.14 kg. A pitcher throws a 40 m/s fastball which is hit by the batter straight back up the middle at a speed of 46 m/s.
a) What is the change in momentum of the ball during the collision with the bat?
b) If this collision occurs during a time of 0.012 seconds, what is the average force exerted by the bat on the ball?

Answer 1

Answer:

(a) The change in momentum is 12.04 kg-m/s

(b) The force exerted by the bat is 1003.33 N

Explanation:

Given that,

The mass of a ball, m = 0.14 kg

Initial speed of the ball, u = 40 m/s

Final speed of the ball, v = -46 m/s

(a) The change in momentum of the ball during the collision with the bat is given by :

$\Delta p=m(v-u)\\\\=0.14(-46-40)\\\\=-12.04\ kg-m/s$

(b) Time for collision, t = 0.012 s

Now the force can be calculated as follows :

$F=\dfrac{\Delta p}{t}\\\\F=\dfrac{12.04}{0.012}\\\\=1003.33\ N$

Hence, this is the required solution.