Two friends, Al and Jo, have a combined mass of 171 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 0.888 m/s, while Jo moves off in the opposite direction at a speed of 1.23 m/s. Assuming that friction is negligible, find Al's mass.
To solve this problem, we can use the principle of conservation of momentum. According to this principle, the total momentum before the release is equal to the total momentum after the release.
The momentum (p) of an object is given by the product of its mass (m) and velocity (v): p = m * v.
Given that Al's velocity after release is 0.888 m/s and Jo's velocity is -1.23 m/s (opposite direction), we can write the momentum equation as:
mass of Al * velocity of Al + mass of Jo * velocity of Jo = 0
We are given that the combined mass of Al and Jo is 171 kg. Let's denote the mass of Al as "m" and the mass of Jo as "171 - m".
Substituting the values into the momentum equation, we get:
m * 0.888 + (171 - m) * (-1.23) = 0
Simplifying the equation:
0.888m - 1.23(171 - m) = 0
0.888m - 210.33 + 1.23m = 0
2.118m = 210.33
m = 210.33 / 2.118
m ≈ 99.283
Therefore, Al's mass is approximately 99.283 kg.