What is the Velocity of current ?

Question

A boat of mass 300 kg was being drifted by the current of the river i.e. the boat was moving at the same speed as the current. A boy with mass of 24 kg was standing still on the boat. Suddenly that boy jumped off the boat in the direction of the current with a velocity of 8ms-1. After that the boat appeared to be at rest for a short moment with respect to a tree on the riverside. What was the velocity of the current?

Answer 1

This problem has to do with conservation of momentum.  The boat and the boy carry a certain amount of momentum with them, but when the boy jumps off the boat, he takes some momentum away from the boat.  But the total amount of momentum in the boy and the boat stay constant at the moment of the jump.  (After the jump, the current exerts a force on the boat and changes the total momentum, but this problem only concerns the moment of the jump.)

 

Momentum (usually referred to as p) equals mass times velocity.

 

So Let v be the velocity of the current relative to the riverside, which is the velocity of the boat and the boy before the jump.

 

The total mass of the boat and the boy is 324 kg, so the total momentum of the boat and the boy together (again relative to the riverside) is 324v.

 

When the boy jumps off, he carries some momentum with him.  How much?  His velocity relative to the boat is 8, so his velocity relative to the riverside is 8+v.  So his momentum is 24(8+v) = 192 + 24v.

 

So the momentum remaining in the boat at that moment is 324v - (192 + 24v) = 300v - 192. 

 

Now, since momentum = mass times velocity, velocity = momentum divided by mass.

 

So since the boat's mass is 300, the velocity of the boat right after the boy jumps off is (300v - 192) / 300 = v - 192/300.

 

But remember, all the velocity and momentum values have been relative to the riverside.  And since the boat appeared to stand still relative to the riverside after the boy jumped, the boat's velocity relative to the riverside at that moment must have been 0.

 

So if v - 192/300 = 0, then v = 192/300 = .64

 

So the velocity of the current was .64 meters/sec

 

 

 

 

Answer 2

The easiest way to solve this problem is to use the principle of conservation of momentum.

In order to use this principle, one must select an inertial frame of reference.   For this problem, the best choice is a frame of reference moving with the water in the stream.     With this choice, the momentum before the jump is zero.  The momentum after the jump is the sum of two terms:  24 kg times 8 m/s =  192 and 300 kg times vb, where vb is the unknown velocity of the boat with respect to the water.   Setting momenta before and after equal results in an equation for  vb ,  namely 192 + 300 vb = 0.   The solution is vb = - 0.64 m/s.   This is negative because the boat recoils from the jump.   The value - 0.64 is relative to the water.  The value relative to the shore tree is - 0.64 + vcurrent, where vcurrent is the velocity of the current with respect to the tree.   Since the value relative to the tree is zero ("boat appeared to be at rest") it is concluded that vsteram = + 0.64