Question

A 121-kg astronaut (including space suit) acquires a speed of 2.20 m/s by pushing off with her legs from a 1600-kg space capsule. Use the reference frame in which the capsule is at rest before the push. Part A: What is the velocity of the space capsule after the push in the reference frame? Express your answer to two significant figures and include the appropriate units. Enter positive value if the direction of the velocity is in the direction of the velocity of the astronaut and negative value if the direction of the velocity is in the direction opposite to the velocity of the astronaut. Part B: If the push lasts 0.600 s , what is the magnitude of the average force exerted by each on the other? Express your answer to three significant figures and include the appropriate units. Part C: What is the kinetic energy of the astronaut after the push in the reference frame? Express your answer to three significant figures and include the appropriate units. Part D What is the kinetic energy of the capsule after the push in the reference frame? Express your answer to two significant figures and include the appropriate units.

Answer 1

Part- A

Solve this part using conservation of momentum.

As we know, momentum is mass times velocity. Initially neither astronaut nor capsule was moving, therefore the total momentum of the system is 0 kg m/s. So the total momentum after the push must be 0 kg m/s also, which implies that the momentum of the capsule must be equal in size (but opposite in direction, however we're only interested in size) to the astronaut's momentum.

Momentum of astronaut = 121 kg * 2.20 m/s

= 266.2 kg m/s.

Therefore, the momentum of the capsule is 266.2 kg m/s in size too.

Now, dividing the momentum by the mass of the capsule (1600 kg) gives the velocity.

So, velocity of space capsule, Vs = 266.2 / 1600 = 0.17 m/s

The direction of velocity of the space capsule is opposite to the direction of astronaut. So, our answer, Vs = -0.17 m/s (Answer)

Part B -

Magnitude of force = Change in momentum divided by time = (266.2 kg m/s)/(0.600 s) = 443.7 N.

= 444.0 N (in three significant digits)

Part C -

Kinetic energy of the astronaut, KEa = 1/2 mv².

= (1/2)*(121*2.20^2) = 292.82 J = 293.0 J (Answer)

Part D -

Kinetic energy of the capsule, KEc = (1/2) * (1600 * 0.17^2) = 23.12 J = 23.0 J (Answer)