Question

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

player who has a mass of 90 kg. Upon impact, the first player immediately stops, thus propelling
the second player backwards. What is the final velocity of the second player?

Answer 1

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

Step-by-step 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)} = (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!