Question

Two friends, Al and Jo, have a combined mass of 170 kg. At the ice skating rink, they stand close together on skates, at rest and facing each other. Using their arms, they push on each other for 1 second and move off in opposite directions. Al moves off with a speed of 11.8 m/sec in one direction and Jo moves off with a speed of 7.6 m/sec in the other. You can assume friction is negligible.

1) What is Al's mass?

2)What is Jo's mass?

3)If you assume the force is constant during the 1 second they are pushing on each other, what is the magnitude of the force of Al on Jo?

4)If you assume the force is constant during the 1 second they are pushing on each other, what is the magnitude of the force of Jo on Al?

Answer 1

Let m be the mass of Al. The mass of Jo is 170 - m
Using the conservation of linear momentum, the momentum before and after the push is the same.
The linear momentum before the push is zero. The linear momentum after the push,
m x vAl + (170 - m) vJo =0
Where v is the velocity. The subscript Al and Jo denotes Al and Jo respectively.
m x 11.8 m/s = (170 - m) x 7.6 m/s
19.4 m =  1292
m = 66.6 kg
Al's mass is 66.6 kg
b) The mass of Jo is
= 170 - 66.6 =  103.4 kg
c) We can find this using the impulse-momentum relation. The impulse J is related to the momentum and the force as
J = F t = p
When Al applies force on Jo, Jo's momentum changes, from 0 to mv, where m is the mass of Jo and v is the final velocity of Jo.
p = m v = 103.4 kg x 7.6 m/s
p = 785.84
The force applied by Al on Jo is
F = p / t
F = 785.84 / 1 s = 785.84 N
D) The force applied by Jo on Al,
p = m v = 785.84 N
F = p / t
F = 785.84 / 1 s = 785.84 N.
The force applied by Jo on Al is same as that of force applied by Al on Jo