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 1.5 m/s 1.5 m/s 4.5 m/s 4.5 m/s 3.0 m/s 3.0 m/s 2.0 m/s
The momentum of an object is defined as its mass multiplied by its velocity. According to the law of conservation of momentum, the total momentum before an event must be equal to the total momentum after the event.
Before the push, both skater A and skater B are at rest, so their velocities are 0 m/s. The total momentum is therefore:
Total momentum before = (Mass of skater A * Velocity of skater A) + (Mass of skater B * Velocity of skater B)
= (72 kg * 0 m/s) + (48 kg * 0 m/s)
= 0 kg*m/s
After the push, skater B has a velocity of 3 m/s in one direction. To prove conservation of momentum, skater A must have an opposite velocity so that their momenta cancel each other out.
Let's say skater A has a velocity of v m/s in the opposite direction. The total momentum after the push is:
Total momentum after = (Mass of skater A * Velocity of skater A) + (Mass of skater B * Velocity of skater B)
= (72 kg * (-v)) + (48 kg * 3 m/s)
= -72v + 144 m/s*kg
To prove conservation of momentum, the total momentum before and after the push must be equal. Therefore:
0 kg*m/s = -72v + 144 m/s*kg
Solving for v, we have:
72v = 144 m/s*kg
v = 144 m/s*kg / 72
v = 2 m/s
Therefore, the velocity of skater A in the opposite direction after the push must be 2 m/s.