Question

in a hockey game, an 80kg player skating at 10 m / s / takes and bumps from behind a 100kg player who is moving the same direction at 8 m / s. as a result of being bumped from behind the 100 kg player speed increases to 9.78 m / s. what is the 80 kg players velocity after the bump​

Answer 1

`7.78\ m/s` is the 80 kg players velocity after the bump.

Step-by-step explanation:

Given the mass and velocity of players.

Velocity before collision is
`10\ m/s\ and\ 8\ m/s` for
`80` kg and
`100` kg player respectively.

And velocity after collision is
`v_1` for
`80` kg player and
`9.78\ m/s` for
`100` kg player

We will apply principle of conservation of momentum.

Momentum
`=(mass)*(velocity)`

So, momentum before collision is
`(80*10)+(100*8)`

And momentum after collision is
`(80* v_1)+(100* 9.78)`

That is momentum before collision should equal momentum after collision.

`(80*10)+(100*8)=(80* v_1)+(100* 9.78)\\\\800+800=80* v_1+978\\1600=80* v_1+978\\1600-978=80* v_1\\622=80* v_1\\(622)/(80)=v_1\\ v_1=7.775`

`v_1=7.775`≅
`7.78\ m/s`