Two ice skaters want to prove conservation of momentum. Skater A has a mass of 72 kg, and skater B has a mass of 48 kg. They are originally at rest. They hold hands and face each other. They then push off each other and move in opposite directions. Skater B moves with a velocity of 3 m/s.
What must the velocity of skater A be in the opposite direction after the push in order to prove conservation of momentum?
(1 point)
Responses
3.0 m/s
3.0 m/s
4.5 m/s
4.5 m/s
2.0 m/s
2.0 m/s
1.5 m/s
To prove conservation of momentum, the total momentum before the push must be equal to the total momentum after the push.
Before the push, both skaters are at rest, so their total momentum is zero.
After the push, skater B moves with a velocity of 3 m/s in the opposite direction. Skater A must have a velocity that cancels out the momentum of skater B.
The momentum of skater B is calculated by multiplying their mass (48 kg) by their velocity (3 m/s):
Momentum of skater B = 48 kg * 3 m/s = 144 kg·m/s
For conservation of momentum, the total momentum after the push must be zero. Therefore, skater A's velocity must be such that their momentum cancels out the momentum of skater B.
Let's assume skater A has a velocity of v m/s. The momentum of skater A is then calculated by multiplying their mass (72 kg) by their velocity (v m/s):
Momentum of skater A = 72 kg * v m/s = 72v kg·m/s
For the total momentum to be zero, the momentum of skater A must be equal in magnitude but opposite in direction to the momentum of skater B. Therefore, we can set up the equation:
72v = -144
To solve for v, divide both sides of the equation by 72:
v = -144 / 72
Simplifying, we find:
v = -2 m/s
Therefore, skater A must have a velocity of -2 m/s in the opposite direction after the push in order to prove conservation of momentum.