A golfer hits a ball with initial velocity 53.5 m/s directed at some angle betwe

Question

A golfer hits a ball with initial velocity 53.5 m/s directed at some angle between 25.0 degrees and 35.0 degrees above the horizontal. The time of contact between the club and ball is 8.0 ms. Take the mass of the golf ball to be 46.0 g and neglect air resistance. (a) What is the magnitude of the average force which the club applied to the ball? (b) By how much (give magnitude only) has the ball's momentum changed 1.3 s after being hit? (a) N (b) kg-m/s (c) What is the direction of the momentum change for Part (b)? Horizontally in direction ball was hit. Horizontally in direction opposite to direction ball was hit. Upward. Downward. Cannot be determined since exact direction in which ball was hit is not given.

Explanation / Answer

a) Average force appllied by club = Fav = change in momentum/time taken

change in momentum p = mv-mu = m(v-u)

m= 46 g = 0.046 kg

u = velocity before contact = 0

v = velocity after contact lost = 53.5 m/s

p = 0.046( 53.5 -0) = 2.461 kg - m/s

t = time of contact = 8ms = 8 * 10-3 s

Fav = 2.461/8 * 10-3 = 307.625 N

b) Note change in momentum is caused by some force now after ball is hit there is no force in horizontal direction

so no change in momentum in horizontal direction , however in vertical direction gravitational force(mg) is acting

so Change in momentum after 1.3 sec = F * t = mg * t = 0.046 * 9.8 * 1.3 = 0.58604 kg-m/s

c) According to 2nd law of motion rate of change of momentum is in direction of force

so change in momentum is also in the direction of force i.e. Downward

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