An avid fan attending a softball game decides to jump the barrier during the thi

Question

An avid fan attending a softball game decides to jump the barrier during the third inning, and run to the pitcher to request an autograph. The pitcher tries to stop the fan by throwing softballs at him. The fan’s mass is 45 kg and he’s running with a speed of 2.88 m/s. The pitcher can throw a 145-gram softball at a speed of 40.0 m/s. What is the minimum number of balls that must hit the fan in order to stop him? (Assume each hit is an inelastic collision.)(Also, these softballs are special, very soft balls, so nobody gets hurt.)( The answer must be an integer.)

Explanation / Answer

Solution:

Mass of the fan = m1= 45 kg

velocity of the fan = v1 = 2.88 m/s

mass of the ball = m =0.145 kg

velocity of the ball = v =-40m/s (negative because the ball and the fan are moving in opposite directions)

Let the velocity of fan +ball after the inelastic collision = V

According to momentum conservation, the momentum of the system before a collision is equal to the momentum of the system after the collision.

so m1v1 -mv = (m1+m)V

=> V=( m1v1 -mv ) /(m1+m)

        = (45 * 2.88 +0.145 * (-40) / (45 + 0.145)

      = 2.74 m/s

when a single ball hits , this is the velocity with which the fan runs towards the pitcher.

(45+0.145) *2.74 = N * 0.145 * 40

=> N = 45.145 * 2.74 / (0.145 * 40 ) = 21

No. of balls = N = 21

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