(a) 0.30 m/s
The total momentum of the rifle+bullet system before the shot is zero:
[tex]p_i = 0[/tex]
The total momentum of the system after the shot is the sum of the momenta of the rifle and of the bullet:
[tex]p_f = m_r v_r + m_b v_b[/tex]
where we have
[tex]m_r = \frac{W}{g}=\frac{35 N}{9.8 m/s^2}=3.57 kg[/tex] is the mass of the rifle
[tex]v_r[/tex] is the final velocity of the rifle
[tex]m_b = 4.5 g = 0.0045 kg[/tex] is the mass of the bullet
[tex]v_b = 240 m/s[/tex] is the final velocity of the bullet
Since the total momentum must be conserved, we have
[tex]p_i = p_f[/tex]
So
[tex]m_r v_r + m_b v_b=0[/tex]
and so we can find the recoil velocity of the rifle:
[tex]v_r = - \frac{m_b v_b}{m_r}=-\frac{(0.0045 kg)(240 m/s)}{3.57 kg}=-0.30 m/s[/tex]
And the negative sign means it travels in the opposite direction to the bullet: so, the recoil speed is 0.30 m/s.
(b) 0.016 m/s
The mass of the man is equal to its weight divided by the acceleration of gravity:
[tex]m=\frac{W}{g}=\frac{650 N}{9.8 m/s^2}=66.3 kg[/tex]
This time, we have to consider the system (man+rifle) - bullet. Again, the total momentum of the system before the shot is zero:
[tex]p_i = 0[/tex]
while the total momentum after the shot is
[tex]p_f = m_r v_r + m_b v_b[/tex]
where this time we have
[tex]m_r = 66.3 kg+3.57 kg=69.9 kg[/tex] is the mass of the rifle+person
[tex]v_r[/tex] is the final velocity of the man+rifle
[tex]m_b = 4.5 g = 0.0045 kg[/tex] is the mass of the bullet
[tex]v_b = 240 m/s[/tex] is the final velocity of the bullet
Since the total momentum must be conserved, we have
[tex]m_r v_r + m_b v_b=0[/tex]
and so we can find the recoil velocity of the man+rifle:
[tex]v_r = - \frac{m_b v_b}{m_r}=-\frac{(0.0045 kg)(240 m/s)}{66.9 kg}=-0.016 m/s[/tex]
So the recoil speed is 0.016 m/s.
