A man jumped at a speed of 5 metres per second from a stationary boat and the boat moved off with the speed of 0.5 metre per second. How many times is the mass of the boat greater than that of the man ?

Apply conservation of linear momentum for the man+boat system (initial momentum = 0). Let m be the man’s mass and M the boat’s mass. After the jump the man’s forward momentum is m·5 and the boat’s recoil momentum is M·0.5 in the opposite direction; total momentum must remain zero, so m·5 = M·0.5. Solving gives M/m = 5/0.5 = 10. Thus the boat’s mass is 10 times the man’s mass. This uses the momentum-conservation principle and straightforward algebraic substitution consistent with standard worked solutions of the problem.