A golfer hits a golf ball of mass 0.045kg the ball over some short trees. he hit

Question

A golfer hits a golf ball of mass 0.045kg the ball over some short trees. he hits the ball at an angle of 60 degreed to the horizontal and it travels a horizontal distance (range) of 60m in a time of 8 s. the golf club of mass 0.60 kg is in contact with the ball for a time of 2.40 milliseconds.

A) What is the average impulsive force of the golf ball?

B) What is the average impulsice force on the golf club?

C) What is the change in momentum of the golf club?

Please show work so I can follow along to learn how to do this. Thanks a bunch!

Explanation / Answer

Taking the horizontal velocity to be .. 60.0m / 8.0s .. Vh = 7.50 m/s The velocity at 60º .. V = 7.50 / cos60 = 15.0 m/s Force = rate of change of momentum = m.?v / t F= (0.045kg x 15.0m/s) / (2.40^-3) .. .. .. ?F = 281.25 N .. at 60º on ball According to Newton law 3 .. The reaction force on the club is equal but opposite in direction .. ?F = - 281.25 N at -60º Mom. change for club is equal and opposite to the mom. change of the ball .. ?mom (ball) = 0.045kg x 15.0m/s = 0.675 kg.m/s ?mom for club .. ?= - 0.675 kg.m/s

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