Question

A hockey player of mass 50kg runs at 20 m/s toward another player of 40kg, moving at -10 m/s. They collide. What are the final velocities of the players?

Answer 1

The final velocity of the players after the collision is `6.7m/s` to the right.

Explanation:

This is a problem of momentum (`vecp`) conservation, where

`vecp=mvecv`,

and because momentum is always conserved, in a collision:

`vecp_f=vecp_i`.

I assume this is meant to be an inelastic collision, where both players share a common final velocity.

We are given that `m_1=50kg`, `v_1=20m/s`, `m_2=40kg`, and `v_2=-10 m/s`

Note that I have defined to the right as the positive direction, and therefore the velocity of the player moving to the left is negative.

The momentum of the first player (`m_1`) before the collision is given by

`vecp_(1)=m_1v_1`

`vecp_1=(50kg)(20m/s)`

`vecp_1=1000(kgm)/s`

Similarly, the momentum of the second player (`m_2`) before the collision is given by

`vecp_2=(40kg)(-10m/s)=-400(kgm)/s`

The total linear momentum before the collision is the sum of the momentums of each of the hockey players.

`vecP=vecp_(t ot)=sumvecp`

`vecP_i=vecp_1+vecp_2`

`vecP_i=1000(kgm)/s+(-400(kgm)/s)=600(kgm)/s`

Because momentum is conserved in all collisions, we know that given a momentum of `600(kgm)/s` before the collision, the momentum after the collision must be `600(kgm)/s`.

After the collision, we have:

`vecP_f=vecp_(1f)+vecp_(2f)`

`vecP_f=m_1v_(f)+m_2v_(f)`

Note that the final velocities must be equal (inelastic collision). We want to solve for `v_f`.

Factor out `v_f`:

`vecP_f=v_f *(m_1+m_2)`

`=>v_f=(vecP_f)/(m_1+m_2)`

Using our known values:

`v_f=(600(kgm)/s)/(50kg+40kg)`

`v_(f)=6.67m/s`

`:.` The final velocity of the players is `6.7m/s` to the right.