A 63 kg astronaut on a spacewalk when his line breaks. He throws a 10 kg o2 tank away from the ship with a v=12 m/s. This propels him back to the ship. Assume his v1=0; what is his v2?
momentum east = momentum west because it was all zero before the toss. :)
63 v2 = 10 * 12
To find the final velocity (v2) of the astronaut, we can use the principle of conservation of momentum. This principle states that the total momentum before an event is equal to the total momentum after the event, provided no external forces are acting on the system.
In this case, we can consider the astronaut and the O2 tank as a system. The momentum before the O2 tank is thrown away is the sum of the momentum of the astronaut and the momentum of the O2 tank, since they are initially at rest:
Initial momentum (before throwing the O2 tank) = 0
After throwing the O2 tank, the sum of the momenta (momentum of the astronaut + momentum of the O2 tank) is conserved:
Final momentum (after throwing the O2 tank) = 0
Mathematically, we can express this as:
(m1 * v1) + (m2 * v2) = 0
Where m1 and m2 are the masses of the astronaut and the O2 tank respectively, and v1 and v2 are their respective velocities.
Substituting the given values, we have:
(63 kg * 0 m/s) + (10 kg * 12 m/s) = 0
This simplifies to:
120 kg * m/s = 0
Since any number multiplied by 0 is 0, we can see that the equation is satisfied. Therefore, the final velocity (v2) of the astronaut is 0 m/s.