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

Answer:

a) V = 5.47m/s

b) $\Delta KE = -188.521J$

Explanation:

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

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

Solving for V:

$V = \frac{mp*Vp+mb*Vb}{mp+mb} =5.47m/s$

The initial kinetic energy was:

$Ki = \frac{mp*Vp^2}{2}+\frac{mb*Vb^2}{2}  =1560.391J$

The final kinetic energy was:

$Kf = \frac{(mp+mb)*V^2}{2}  =1371.87J$

So, the change in the kinetic energy was:

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