Answer:
25 m/s in the opposite direction with the ship recoil velocity.
Explanation:
Assume the ship recoil velocity and velocity of the cannon ball aligns. By the law of momentum conservation, the momentum is conserved before and after the shooting. Before the shooting, the total momentum is 0 due to system is at rest. Therefore, the total momentum after the shooting must also be 0:
[tex]m_sv_s + m_bv_b = 0[/tex]
where [tex]m_s = 2500 kg, m_b = 25 kg[/tex] are masses of the ship and ball respectively. [tex]v_s = 0.25 m/s, v_b[/tex] are the velocities of the ship and ball respectively, after the shooting.
[tex]2500*0.25 + 25*v_b = 0[/tex]
[tex]25v_b = -2500*0.25[/tex]
[tex]v_b = -2500*0.25/25 = -25 m/s[/tex]
So the cannon ball has a velocity of 25 m/s in the opposite direction with the ship recoil velocity.