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

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lcmendozaf lcmendozaf · Answer 1 · 2019-10-30T22:40:38+01:00

Answer:

a) V = 5.47m/s

b) [tex]\Delta KE = -188.521J[/tex]

Explanation:

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

[tex]mp*Vp + mb*Vb = (mp+mb)*V[/tex]

Solving for V:

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

The initial kinetic energy was:

[tex]Ki = \frac{mp*Vp^2}{2}+\frac{mb*Vb^2}{2} =1560.391J[/tex]

The final kinetic energy was:

[tex]Kf = \frac{(mp+mb)*V^2}{2} =1371.87J[/tex]

So, the change in the kinetic energy was:

[tex]\Delta KE = Kf - Ki=-188.521J[/tex]