# Math Help - find the mass of one of them.

1. ## find the mass of one of them.

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 0.95 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.

in kg

must i use an equation for tension, or is that relevant in this problem?

2. Originally Posted by rcmango
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 0.95 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.

in kg

must i use an equation for tension, or is that relevant in this problem?
I suppose you could say that there is a "tension" in the spring, but I wouldn't worry about it.

There is no net external force on the two skaters. (Yes, there is a force on each by the spring, but this is internal to the skater-skater-spring system.) Thus the total momentum of the system is conserved:
$P_{0, tot} = P_{tot}$

I'm going to define a coordinate system such that Jo is in the +x direction compared to Al. (In other words, the +x axis is in the direction Jo is going to be moving off in after the spring is released.)

Before the release of the spring the total momentum of the system was 0 kg m/s. So
$P_{tot} = 0$

$m_{Al}v_{Al} + m_{Jo}v_{Jo} = 0$

$m_{Al} \cdot (-0.95) + (168 - m_{Al}) \cdot 1.1 = 0$
(Remember this is a vector equation, and Al moves in the -x direction. So his velocity has a negative component in the x direction.)

You can solve this for $m_{Al}$. For reference I get $m_{Al} = 90.1463~kg~m/s$.

-Dan