A)A 0.145 kg ball is dropped fromrest. If the magnitude of the ball\'s momentum

Question

A)A 0.145 kg ball is dropped fromrest. If the magnitude of the ball's momentum is 0.760 kg·m/s just beforeit lands on the ground, from what height was it dropped? B)In a typical golf swing, the club is in contact with theball for about 0.0011 s. If the 45 gball acquires a speed of 71 m/s,estimate the magnitude of the force exerted by the club on theball.
A)A 0.145 kg ball is dropped fromrest. If the magnitude of the ball's momentum is 0.760 kg·m/s just beforeit lands on the ground, from what height was it dropped? B)In a typical golf swing, the club is in contact with theball for about 0.0011 s. If the 45 gball acquires a speed of 71 m/s,estimate the magnitude of the force exerted by the club on theball.

Explanation / Answer

a) Let the ball is dropped from a height 'h'm     Velociti acquirde by the ball on reachingthe ground, v=2gh ...(1)    Momentum acquired by the ball on reaching theground, P=mv =0.760kg.m/s                                                             Mass of the ball ,m=0.145kg                                                        Velocity of the ball,v =P/m=0.760/0.145=5.24m/s           Therefore, height from which ball is driopped,h= v2 /2g=(5.24)2 /2x9.8= 1.4m b) Mass of the ball, m= 45g or 45x10-3 kg    Speed acquired, v= 71m/s     Time of contact with the ball,t=0.0011s Mometum attained by the ball, P=mv Inital momentum=0 Now, force, F= Rate of change of momentum=mv/t=45x10-3 /0.0011=40.90 N Magnitude of force excerted on the ball,F=40.90N    Hope this may help you

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