Question

A 91.5 kg football player running east at 2.73 m/s tackles a 63.5 kg player running east at 3.09 m/s. what is their velocity afterward? PLEASE HELP

Answer 1

Velocity of the players afterwards = 2.88 m/s towards east.

Step-by-step explanation:

Mass of football player A
`(m_1)`= 91.5 kg

Velocity of player A
`(v_1)`= 2.73 m/s

Mass of football player B
`(m_2)` =63.5 kg

Velocity of player B
`(v_2)`= 3.09 m/s

Since both players move in same direction east, so their velocity afterwards will also be in same direction east.

By law of conservation of momentum we have:

Momentum before collision = Momentum after collision

This can be written as:

`p_i=p_f`

`m_1v_1+m_2v_2=(m_1+m_2)\ v`

where
`v` is the velocity of the players together afterwards.

We can plugin the given value to find
`v`

`(91.5)(2.73)+(63.5)(3.09)=(91.5+63.5)\ v`

`249.795+196.215=155\ v`

`446.01=155\ v`

Dividing both sides by 155.

`(446.01)/(155)=(155\ v)/(155)`

`2.88=v`

∴
`v=2.88` m/s

Velocity of the players afterwards = 2.88 m/s towards east.