Question

A cricket ball has mass 0.155 kg. If the velocity of a bowled ball has a magnitude of 35.0 m/s and the batted ball's velocity is 65.0 m/s in the opposite direction, find the magnitude of the change in momentum of the ball.
Find the magnitude of the impulse applied to it by the bat.
If the ball remains in contact with the bat for 2.00 ms , find the magnitude of the average force applied by the bat.

Answer 1

Answer:

Magnitude of change in momentum = 4.65 kg.m/s

Magnitude of impulse = 4.65 kg.m/s

Magnitude of the average force applied by the bat = 1550 N

Explanation:

Mass of the cricket ball, m = 0.155 kg

Initial velocity of the ball, u = 35.0 m/s

final velocity of the ball after hitting the bat, v = 65.0 m/s

Time of contact, t = 2.00 ms = 2.00 × 10⁻³ s

Now,

Magnitude of change in momentum = Final momentum - Initial momentum

or

Magnitude of change in momentum = ( m × v ) - ( m × u )

or

Magnitude of change in momentum = ( 0.155 × 65 ) - ( 0.155 × 35 )

or

Magnitude of change in momentum = 10.075 - 5.425 = 4.65 kg.m/s

Now, Magnitude of impulse = change in momentum

thus,

Magnitude of impulse = 4.65 kg.m/s

Now,

magnitude of the average force applied by the bat = $\frac{\textup{Impulse}}{\textup{Time}}$

or

magnitude of the average force applied by the bat = $\frac{\textup{4.65}}{\textup{3}\times\textup{10}^{-3}}$

or

Magnitude of the average force applied by the bat = 1550 N