Two ice skaters stand at rest in the center of an ice rink. When they push off against one another the 62-kg skater acquires a speed of 0.70 m/s. If the speed of the other skater is 0.90 m/s, what is this skater's mass?

48.22 kg

Explanation:

Applying the law of conservation of momentum,

Total momentum before collision = Total momentum after collision

Note:  Since both skaters were initially at rest, then their total momentum  before collision is equal to zero.

And the velocity of the second skater will be in opposite direction to the first.

0 = mv+m'v' Equation 1

Where m = mass of the first skater, m' = mass of the second skater, v = final velocity of the first skater, v' = final velocity of the second skater.

make v' the subject of the equation

m' = -mv/v' Equation 2

Given: m = 62 kg, v = 0.7 m/s, v' = -0.9 m/s (opposite direction to the first)

Substitute into equation 1

m' = -62(0.7)/-0.9

m' = 48.22 kg

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