Two ice skaters want to prove the conservation of momentum. Skater A has a mass of 72 kg and skater B has a mass of 55 kg. They are originally at rest. They hold hands and face each other

Answer 1

To prove the conservation of momentum, we need to show that the total momentum before and after the event remains the same.

Before they hold hands, both skaters are originally at rest, so their initial momenta are zero.

The momentum of an object is given by the equation:

momentum = mass × velocity

Let's assume that after they hold hands, they push each other away, and skater A moves to the right with a velocity of v1 m/s, and skater B moves to the left with a velocity of v2 m/s.

The total momentum before the event is the sum of the individual momenta of skater A and skater B, both of which are zero:

Total initial momentum = momentum of skater A + momentum of skater B
= 0 + 0
= 0

After the event, the total momentum should still be zero for the conservation of momentum to hold. Therefore:

Total final momentum = momentum of skater A + momentum of skater B
= (mass of skater A × velocity of skater A) + (mass of skater B × velocity of skater B)
= (72 kg × v1 m/s) + (55 kg × v2 m/s)

To maintain the conservation of momentum, skater A pushes skater B with the same force he receives from skater B, just in the opposite direction. Therefore, the magnitudes of their velocities will be equal:

v1 = -v2

Thus, the total final momentum equation becomes:

Total final momentum = (72 kg × v1) + (55 kg × (-v1))
= 72v1 - 55v1
= 17v1

To satisfy the conservation of momentum, the total final momentum should be zero:

17v1 = 0
v1 = 0

Hence, the velocity of skater A, v1, is zero, which means skater A does not move.

Since skater A doesn't move, the velocity of skater B, v2, is also zero, so skater B does not move either.

In conclusion, when skater A and skater B hold hands and face each other, they do not move, and the conservation of momentum is proven.