In CSU Rams football game ( American fotball, no soccer) an 80kg CSU receiver ju
ID: 1538987 • Letter: I
Question
In CSU Rams football game ( American fotball, no soccer) an 80kg CSU receiver jumps straight in the air and catches a 0.5kg football thrown towards him. just before catch. the receiver is a height 0.5m off the ground. and the football is travelling horizontally at a speed of 40m/s (90mph).
assuming the catch is perfectly inelastic collision (the ball and the layer move togather after-wards):
1 how fast are the ball and the player moving immeditely after catch?
2 what is the total kinetic energy immeditely after catch?
3 how much energy was lost during the collision?
Explanation / Answer
Given
mass of foot ball m = 0.5 kg
mass of receiver M = 80 kg
speed of the ball before catch is V1 = 10 m/s
1) conservation of momentum m1v1+m2v2 = (m1+m2)V
0.5*10 +80*0 = (80.5)v
V2 = 0.062112 m/s
the ball and player will move with speed V2 = 0.062112 m/s
2) total k.e immediately after catch is
initial k.e = 0.5*mV1^2 = 0.5*0.5*10^2 = 25 J
finall k.e after catch = 0.5*(M+m)v2^2 = 0.5(80.5)(0.062112)^2 J = 0.155280 J
k.e immediately after catch is = 0.155280 J
3) energy difference is k2-k1 = 0.155280 - 25 J = -24.84472 J
so the energy losss is -24.84472 J