Two hockey players traveling at 12 m/s (Player #1) and 19 m/s (Player #2) hit in a head-on collision. They grab onto each other to keep from falling and slide in the same direction Player #2 was traveling at a speed of 5 m/s. Player #1 has a mass of 100 kg. What is Player #2’s mass?
So far I've tried to use the momentum equation:
m1v1 + m2v2 = (m1+m2) v'
(100)(12) + m2(-19) = (100 + m2) (-5)
But I keep getting the wrong answer based off my teacher's answer key.
It seems like you are on the right track by using the principle of conservation of momentum. However, there appears to be a sign error in your equation.
The correct equation to use is:
m1v1 + m2v2 = (m1 + m2)v'
Here, m1 is the mass of Player #1, v1 is the initial velocity of Player #1, m2 is the mass of Player #2, v2 is the initial velocity of Player #2, and v' is the final velocity at which the two players are sliding.
Plugging in the given values:
(100 kg)(12 m/s) + m2(-19 m/s) = (100 kg + m2)(-5 m/s)
Simplifying the equation further, we get:
1200 kg·m/s - 19m2 = -500 kg·m/s - 5m2
To solve for m2, let's isolate the m2 terms on one side:
-19m2 + 5m2 = -500 kg·m/s - 1200 kg·m/s
Combine like terms:
-14m2 = -1700 kg·m/s
Dividing both sides of the equation by -14, we get:
m2 = 1700 kg·m/s / 14
Calculating this, we find:
m2 ≈ 121.43 kg
Therefore, Player #2's mass is approximately 121.43 kg.