Question

Consult Concept Simulation 7.1 in preparation for this problem. Two friends, Al and Jo, have a combined mass of 132 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 1.04 m/s, while Jo moves off in the opposite direction at a speed of 0.784 m/s. Assuming that friction is negligible, find Al's mass.

Answer 1

Answer:

56.7 kg

Explanation:

We can solve this problem by using the law of conservation of momentum. In fact, the total momentum at beginning must be equal to the total momentum after the move far apart from each other.

Since Al and Jo are at rest at the beginning, their total momentum is zero:

p = 0 (1)

After the release their arms, the total momentum is:

$p=m_A v_A + m_J v_J$ (2)

where

$m_A$ is Al's mass

$m_J$ is Jo's mass

$v_A = 1.04 m/s$ is Al's velocity

$v_J = -0.784 m/s$ is Jo's velocity (negative since Jo moves in the opposite direction)

Since momentum must be conserved, we can equate (1) and (2)

$0=m_A v_A + m_J v_J$ (3)

Also, we know that the combined mass of Al + Jo is

$M=m_A + m_J = 132$

So we can rewrite Jo's mass as

$m_J = M-m_A$

And substituting into (3) and re-arranging the equation, we can find Al's mass:

$0=m_A v_A + (M-m_A)v_J\\0=m_A v_A + M v_J - m_A v_J\\m_A = \frac{Mv_J}{v_J-v_A}=\frac{(132)(-0.784)}{-0.784-1.04}=56.7 kg$