**Motion of the boat:**

(i) A man having weighs 500 N and is in a boat weighing 1500 N which is free to move over a steady lake. If he begins running along the length of the boat at a speed of 4 metres/second with respect to the boat-floor, supposing that there is no resistance of water to the motion of the boat, find out the velocity of the boat.

(ii) Then if he jumps over the boat and absolute velocity of 6 metres/seconds, then what would be the motion of the boat?

**Solution**

(i) As both man and the boat are initially at rest, the initial total-momentum is zero. Therefore by the principle of conservation of momentum, the total momentum for this system shall be zero at any other time. If the velocity of man is V_{m/b }with respect to the floor of the boat along x direction, let V_{b} be the absolute velocity of the boat developed due to the interaction developed.

The absolute (or true/actual) velocity of the man V_{m} is, thus, equal to (V_{m/b }+ V_{b} ) . The equation of conservation of momentum is associated with masses m_{b} and mm of the boat and the man, respectively.

∴ m_{b} (V_{b }) + m_{m} (V_{m} ) = 0

(1500/9.8) (V ) +( 500/9.8) (V _{m/b} + V _{b} ) = 0

∴ V_{b} (1500 + 500) = - V_{m/b }× (500)

∴ (V_{b} ) = - (4 × 500) /2000

= - 1 m / sec

Therefore, the boat moves with a velocity of 1 m/sec in negative x direction and true (absolute) velocity of man reduces.

Actual (Absolute) velocity of man = V_{m/b }+ V_{b}

= 4 - 1 = 3 m / sec.

(ii) While the man jumps out of the boat with absolute velocity V_{m}′ = 6 m / sec, the boat shall attain a velocity V_{b}′ whereby,

m_{b} (V_{b}′) + ( m_{m }) V_{m}′ = 0

∴ (1500/9.8) × V _{b}′ = - (500 /9.8) × 6

∴ V_{b}′ = - 2 metres / sec.

Negative sign again indicates that velocity (V_{b}′) of the boat has direction opposite to the direction of the man's jump.