Two people stand facing each other at a roller skating rink and then push each other. If one skater has a mass of 44 kg and moves backward at 1.60 m/s, what is the velocity of the second skater if his mass is 67 kg?

Question

High School

Answer :

DamianDaniel DamianDaniel · Answer 1 · 2024-03-14T18:06:28-04:00

The velocity of the second skater is -1.05 m/s.

The velocity of the second skater can be calculated using the principle of conservation of linear momentum.

The initial momentum of the first skater is given by mass1 * velocity1 = (44 kg) * (-1.60 m/s) = -70.4 kg·m/s (since the direction is backward).

According to the conservation of linear momentum, the total initial momentum of the system should be equal to the total final momentum of the system.

Therefore, the total initial momentum of the system is equal to the total final momentum of the system.

Since there are only two skaters in this system, the final momentum of the first skater is equal to the initial momentum of the second skater.

So, the final momentum of the second skater can be calculated as follows:

mass2 * velocity2 = (67 kg) * velocity2

Therefore, -70.4 kg·m/s = (67 kg) * velocity2

Solving for velocity2

we get velocity2 = -70.4 kg·m/s / (67 kg) = -1.05 m/s.

Therefore, the velocity of the second skater is -1.05 m/s.