(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):
![p_i = 0](https://tex.z-dn.net/?f=p_i%20%3D%200)
The final total momentum is instead:
![p_f = m_a v_a + m_c v_c](https://tex.z-dn.net/?f=p_f%20%3D%20m_a%20v_a%20%2B%20m_c%20v_c)
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
![p_i = p_f = 0](https://tex.z-dn.net/?f=p_i%20%3D%20p_f%20%3D%200)
so
![m_a v_a + m_c v_c=0](https://tex.z-dn.net/?f=m_a%20v_a%20%2B%20m_c%20v_c%3D0)
Solving the equation for
, we find
![v_c = - \frac{m_a v_a}{m_c}=-\frac{(125 kg)(2.50 m/s)}{1900 kg}=-0.165 m/s](https://tex.z-dn.net/?f=v_c%20%3D%20-%20%5Cfrac%7Bm_a%20v_a%7D%7Bm_c%7D%3D-%5Cfrac%7B%28125%20kg%29%282.50%20m%2Fs%29%7D%7B1900%20kg%7D%3D-0.165%20m%2Fs)
(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:
![F \Delta t = \Delta p](https://tex.z-dn.net/?f=F%20%5CDelta%20t%20%3D%20%5CDelta%20p)
The change in momentum of the astronaut is
![\Delta p= m\Delta v = (125 kg)(2.50 m/s)=312.5 kg m/s](https://tex.z-dn.net/?f=%5CDelta%20p%3D%20m%5CDelta%20v%20%3D%20%28125%20kg%29%282.50%20m%2Fs%29%3D312.5%20kg%20m%2Fs)
And the duration of the push is
![\Delta t = 0.600 s](https://tex.z-dn.net/?f=%5CDelta%20t%20%3D%200.600%20s)
So re-arranging the equation we find the average force exerted by the capsule on the astronaut:
![F=\frac{\Delta p}{\Delta t}=\frac{312.5 kg m/s}{0.600 s}=520.8 N](https://tex.z-dn.net/?f=F%3D%5Cfrac%7B%5CDelta%20p%7D%7B%5CDelta%20t%7D%3D%5Cfrac%7B312.5%20kg%20m%2Fs%7D%7B0.600%20s%7D%3D520.8%20N)
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:
![K=\frac{1}{2}mv^2](https://tex.z-dn.net/?f=K%3D%5Cfrac%7B1%7D%7B2%7Dmv%5E2)
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
![K=\frac{1}{2}(125 kg)(2.50 m/s)^2=390.6 J](https://tex.z-dn.net/?f=K%3D%5Cfrac%7B1%7D%7B2%7D%28125%20kg%29%282.50%20m%2Fs%29%5E2%3D390.6%20J)
For the capsule, m = 1900 kg and v = 0.165 m/s, so its kinetic energy is
![K=\frac{1}{2}(1900 kg)(0.165 m/s)^2=25.9 J](https://tex.z-dn.net/?f=K%3D%5Cfrac%7B1%7D%7B2%7D%281900%20kg%29%280.165%20m%2Fs%29%5E2%3D25.9%20J)