A125-kg astronaut (including space suit) acquires a speed of 2.50 m/s by pushing off with her legs from a 1900-kg space capsule. (a) what is the change in speed of the space capsule? (b) if the push lasts 0.600 s, what is the average force exerted by each on the other? as the reference frame, use the position of the capsule before the push. (c) what is the kinetic energy of each after the push?

(a) 0.165 m/s

The total initial momentum of the astronaut+capsule system is zero (assuming they are both at rest, if we use the reference frame of the capsule):

The final total momentum is instead:

where

is the mass of the astronaut

is the velocity of the astronaut

is the mass of the capsule

is the velocity of the capsule

Since the total momentum must be conserved, we have

so

Solving the equation for , we find

(negative direction means opposite to the astronaut)

So, the change in speed of the capsule is 0.165 m/s.

(b) 520.8 N

We can calculate the average force exerted by the capsule on the man by using the impulse theorem, which states that the product between the average force and the time of the collision is equal to the change in momentum of the astronaut:

The change in momentum of the astronaut is

And the duration of the push is

So re-arranging the equation we find the average force exerted by the capsule on the astronaut:

And according to Newton's third law, the astronaut exerts an equal and opposite force on the capsule.

(c) 25.9 J, 390.6 J

The kinetic energy of an object is given by:

where

m is the mass

v is the speed

For the astronaut, m = 125 kg and v = 2.50 m/s, so its kinetic energy is

For the capsule, m = 1900 kg and v = 0.165 m/s, so its kinetic energy is

answer: electrons

answer: 2×10^-2kg

explanation: density= mass per volume

vol=mass/density

converting to s.i units we have;

3÷1000kg = 0.003

1.5÷100m = 0.15

0.003/0.15 = 0.02kg