Question

A 110 kg football player running with a velocity of 5.0 m/s hits another stationary football

Answer 1

Answer:

The final velocity of the second player is 6.1 m/s.

Explanation:

The final velocity of the second player can be calculated by conservation of linear momentum (p):

$p_{i} = p_{f}$  

$m_{a}v_{a_{i}} + m_{b}v_{b_{i}} = m_{a}v_{a_{f}} + m_{b}v_{b_{f}}$  (1)

Where:

$m_{a}$: is the mass of the first football player = 110 kg

$m_{b}$: is the mass of the second football player = 90 kg

$v_{a_{i}}$: is the initial velocity of the first football player = 5.0 m/s

$v_{b_{i}}$: is the initial velocity of the second football player = 0 (he is at rest)

$v_{a_{f}}$: is the final velocity of the first football player = 0 (he stops after the impact)

$v_{b_{f}}$: is the final velocity of the second football player =?

By solving equation (1) for $v_{b_{f}}$ we have:

$110 kg*5.0 m/s + 0 = 0 + 90 kg*v_{b_{f}}$

$v_{b_{f}} = \frac{110 kg*5.0 m/s}{90 kg} = 6.1 m/s$

Therefore, the final velocity of the second player is 6.1 m/s.

I hope it helps you!