Question

When a 0.15 kg baseball is hit, its velocity changes from +17 m/s to -17 m/s.

(a) What is the magnitude of the impulse delivered by the bat to the ball?
N·s

(b) If the baseball is in contact with the bat for 1.5 ms, what is the magnitude of the average force exerted by the bat on the ball?

Answer 1

Explanation:

It is given that,

Mass of the baseball, m = 0.15 kg

Initial velocity, u = +17 m/s

Final velocity, v = -17 m/s

(a) We need to find the magnitude of impulse delivered by the bat to the ball. We know that impulse is equal to the change in momentum of an object i.e.

$J=\Delta p=m(v-u)$

$J=0.15\ kg\times (-17\ m/s-17\ m/s)$

J = −5.1 Kg-m/s

(b) If the baseball is in contact with the bat for 1.5 ms i.e. the time of collision Δt = 0.0015 s

We need to find the magnitude of the average force exerted by the bat on the ball.

$J=F.\Delta t$

$F=\dfrac{J}{\Delta t}$

$F=\dfrac{5.1\ Kg-m/s}{0.0015\ s}$

F = 3400 N

Hence, this is the required solution.