Question

Explosion

1) Two swimmers are floating on a raft that is motionless. One swimmer has a mass of 50 kg and
the other at 80 kg. They both push off the raft at the same time. The 80 kg swimmer moves
away at 3 m/s. What velocity does the 50 kg swimmer move away with?
M1 = 50 kg
M2 = 80 kg
v2' = 3 m/s
Equation: 0= m1 (v1') + m2 (v2')
v1' =
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Answer 1

We can use the law of conservation of momentum to solve this problem, which states that the total momentum of a system before an event (in this case, pushing off the raft) is equal to the total momentum of the system after the event, assuming no external forces act on the system.

Before the push-off, the raft and the two swimmers have a total momentum of zero, since they are motionless. After the push-off, the total momentum of the system is still zero.

Let's assume that the 50 kg swimmer moves away from the raft with a velocity of v1' after the push-off. The 80 kg swimmer moves away from the raft with a velocity of 3 m/s, in the opposite direction.

Using the law of conservation of momentum, we can write:

0 = m1 * v1' + m2 * v2'

where m1 is the mass of the 50 kg swimmer, m2 is the mass of the 80 kg swimmer, and v2' is the velocity of the 80 kg swimmer after the push-off.

Substituting the values given in the problem, we get:

0 = 50 kg * v1' + 80 kg * (-3 m/s)

Solving for v1', we get:

v1' = (80 kg * 3 m/s) / 50 kg

v1' = 4.8 m/s

Therefore, the 50 kg swimmer moves away from the raft with a velocity of 4.8 m/s after the push-off.