Question

Two friends, Al and Jo, have a combined mass of 168 kg. At an ice skating rink they stand close together on skates, at rest and facing each other, with a compressed spring between them. The spring is kept from pushing them apart because they are holding each other. When they release their arms, Al moves off in one direction at a speed of 0.91 m/s, while Jo moves off in the opposite direction at a speed of 1.1 m/s. Assuming that friction is negligible, find Al's mass.

Answer 1

Answer: 91.94 kg

Explanation: Supoose Al's mass is m. The combined mass is 168kg, so Jo's mass is 168 - m.

The Law of Conservation of Momentum states that when two objects collide, the total momentum of both objects before the collision is the same as the total momentum of both objects after the collision.

At the beginning, Al and Jo are united without movement, so their initial momentum is zero.

After the release of their hands, Al goes in one direction and Jo moves to the opposite direction. Suppose the direction Al is moving is positive. Conservation of momentum will be

$m_{1}v_{1}_{(f)}+m_{2}v_{2}_{(f)}=m_{1}v_{1}_{(i)}+m_{2}v_{2}_{(i)}$

where

index i referes to initial momentum

index f to final momentum

index 1 refers to Al

index 2 to Jo

Calculating:

$m(0.91)-[(168-m)(1.1)]=0$

$0.91m+1.1m-184.8=0$

$2.01m=184.8$

m = 91.94

Al has a mass of 91.94 kg.