A golfer hits a golf ball, imparting to it an initial velocity of magnitude 52.2 m/s directed 30° above the horizontal. Assuming that the mass of the ball is 46.0 g and the club and ball are in contact for 1.20 ms, find (a) the impulse imparted to the ball, (b) the impulse imparted to the club, and (c) the average force exerted on the ball by the club.

Question

Navjyot Kalra · Answer

The initial momentum pi of the ball will be,
p_i = mv_i
To obtain the initial momentum p_i of the ball, substitute 46.0 g for the mass of the ball and 52.2 m/s for vi in the equation p_i = mv_i,
p_i = mv_i
= (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 p_f of the ball will be zero.
So, p_f = 0

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