Solution:-
The initial momentum pi of the ball will be,

pi = mvi

To obtain the initial momentum pi of the ball, substitute 46.0 g for the mass of the ball and 52.2 m/s for vi in the equation pi = mvi,

pi = mvi

= (46.0 g) (52.2 m/s)

= (46.0 g×10-3 kg/1 g) (52.2 m/s)

= 2.4 kg.m/s

= (2.4 kg.m/s) (1 N/1 kg.m/s2)

= 2.4 N.s

As the final speed of the ball is zero, thus the final momentum pf of the ball will be zero.

So, pf = 0

To find out the impulse J imparted to the ball, substitute 2.4 N.s for pi and 0 for pf

n the equation J = pi – pf,

J = Δp

= pi – pf

= 2.4 N.s -0

= 2.4 N.s

Therefore the impulse J imparted to the ball would be 2.4 N.s.

b) The impulse J imparted to the club is just opposite that imparted to the ball. Therefore the impulse J imparted to the club will be 2.4 N.s.

c) To obtain the average force F exerted on the ball by the club, substitute 2.4 N.s for Δp and 1.20 ms for t in the equation F = Δp/ Δt,

F = Δp/ Δt

= 2.4 N.s/1.20 ms

= (2.4 N.s)/((1.20 ms)(10-3 s/1ms))

= 2000 N

From the above observation we conclude that, the average force F exerted on the ball by the club would be 2000 N