Question

A 91 kg football player catches a 0.700 kg ball with his feet off the ground with both of them moving horizontally. The player's speed is 5.30 m/s, and the ball's speed is 28.4 m/s. (a) First, consider the situation where the player and the ball are going in the same direction, and take this direction as positive. Calculate their final velocity (in m/s). m/s (5 attempts remaining) Input 1 Status (b) Calculate the change in the kinetic energy (in Joules) of the system (of the player and the ball). ΔKE

Answer 1

a) V = 5.47m/s

b)
`\Delta KE = -188.521J`

Step-by-step explanation:

Since the ball and the player become a whole system after the catch:

`mp*Vp + mb*Vb = (mp+mb)*V`

Solving for V:

`V = (mp*Vp+mb*Vb)/(mp+mb) =5.47m/s`

The initial kinetic energy was:

`Ki = (mp*Vp^2)/(2)+(mb*Vb^2)/(2)  =1560.391J`

The final kinetic energy was:

`Kf = ((mp+mb)*V^2)/(2)  =1371.87J`

So, the change in the kinetic energy was:

`\Delta KE = Kf - Ki=-188.521J`