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?

1 Answer
Dec 21, 2016

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.