Two hockey players are traveling at velocities of v1 = 15 m/s and v2 = -16 m/s when they undergo a head-on collision. After the collision, they grab each other and slide away together with a velocity of -3.8 m/s. Hockey player 1 has a mass of 121 kg. What is the mass of the other player?
keep getting the wrong answer..
Show us your work and someone will tell you where you went wrong.
This is a conservation of momentum problem.
P= mv1i + mv2i = (m1 + m2)vf
= 121kg(15m/s) +m2(-16) = (121+m2)(-3.8)
and I keep going wrong from here..
nvm, got it!
To solve this problem, we can use the principle of conservation of linear momentum. According to this principle, the total momentum before the collision should be equal to the total momentum after the collision.
We can start by calculating the initial momentum of each hockey player. The momentum (p) of an object is defined as the product of its mass (m) and velocity (v): p = m * v.
For Hockey player 1:
Initial momentum (p1) = mass1 * velocity1 = 121 kg * 15 m/s
For Hockey player 2:
Initial momentum (p2) = mass2 * velocity2 = mass2 * (-16 m/s)
Since the collision is head-on, the total initial momentum is the sum of the individual momenta: p_initial = p1 + p2.
Now let's calculate the final momentum after the collision. The hockey players grab each other and slide away with a velocity of -3.8 m/s.
Final momentum (p_final) = (mass1 + mass2) * (-3.8 m/s)
According to the principle of conservation of linear momentum, p_initial = p_final.
Therefore, we have the following equation:
p1 + p2 = (mass1 + mass2) * (-3.8 m/s)
Substituting the values we know:
121 kg * 15 m/s + mass2 * (-16 m/s) = (121 kg + mass2) * (-3.8 m/s)
Simplifying:
1815 - 16 * mass2 = -459.8 - 3.8 * mass2
Rearranging the equation:
-16 * mass2 + 3.8 * mass2 = -459.8 - 1815
-12.2 * mass2 = -2274.8
Dividing both sides by -12.2:
mass2 = (-2274.8) / (-12.2)
Calculating the value:
mass2 ≈ 186.72 kg
Therefore, the mass of the other hockey player is approximately 186.72 kg.