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?

Question

William Audy · Accepted Answer

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

This is a problem of momentum (#vecp#) conservation, where #vecp=mvecv#, additionally, in a collision, because momentum is always conserved: #vecp_f=vecp_i#. This seems to be intended to be an inelastic collision in which the players have the same ultimate velocity. We are given that #m_1=50kg#, #v_1=20m/s#, #m_2=40kg#, and #v_2=-10 m/s# Because I defined to the right as the positive direction, the player moving to the left is travelling at a negative velocity. 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 sum of the linear momentums of the hockey players before the collision is the total linear momentum. #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#. Following the impact, 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 the values we are aware of: #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.

Nora Arzate · Answer

undefined To solve this problem, we can use the principle of conservation of momentum. The total momentum before the collision is equal to the total momentum after the collision.

Let's denote the initial velocities of the players as follows:
Player 1 (50 kg) runs at 20 m/s
Player 2 (40 kg) moves at -10 m/s (negative indicates moving in the opposite direction)

First, calculate the total momentum before the collision:
Total momentum before collision = (mass of player 1 * velocity of player 1) + (mass of player 2 * velocity of player 2)

Total momentum before collision = (50 kg * 20 m/s) + (40 kg * -10 m/s) = 1000 kg m/s - 400 kg m/s = 600 kg m/s

Since momentum is conserved, the total momentum after the collision is also 600 kg m/s.

Let's denote the final velocities of the players as v1 and v2. Using the conservation of momentum:

Total momentum after collision = (mass of player 1 * final velocity of player 1) + (mass of player 2 * final velocity of player 2)

600 kg m/s = (50 kg * v1) + (40 kg * v2)

Now, we also have the equation for conservation of kinetic energy, since this is an elastic collision:

0.5 * m1 * (initial velocity of player 1)^2 + 0.5 * m2 * (initial velocity of player 2)^2 = 0.5 * m1 * (final velocity of player 1)^2 + 0.5 * m2 * (final velocity of player 2)^2

Solving these two equations simultaneously will give us the final velocities of the players. After solving, the final velocities are approximately:

v1 ≈ 3.08 m/s
v2 ≈ -7.69 m/s