Complete question:
A 0.144 kg baseball approaches a batter with a speed of 20m/s . The batter lines the ball directly back to the pitcher with a speed of 30m/s . Find the impulse exerted on the ball . If the bat and ball were in contact for 0.012 sec , find the average force exerted on the ball by the bat
Answer:
(1) The impulse exerted by the ball is 7.2 kg.m/s
(2) The average force exerted on the ball by the bat is 600 N.
Explanation:
Given;
mass of the baseball, m = 0.144 kg
velocity of the baseball, v₁ = 20 m/s
velocity of the batter, v₂ = -30 m/s (opposite direction to the ball's speed)
The impulse exerted by the ball is calculated as follows;
J = ΔP = mv₁ - mv₂
ΔP = m(v₁ - v₂)
ΔP = 0.144 [20 - (-30)]
ΔP = 0.144 ( 20 + 30 )
ΔP = 0.144 (50)
ΔP = 7.2 kg.m/s
The average force exerted on the ball by the bat is calculated as;
[tex]F = \frac{\Delta P }{t} \\\\where; t \ is \ time \ of \ contact= 0.012 \ s\\\\F = \frac{7.2}{0.012} \\\\F = 600 \ N[/tex]
