Question

boy of mass 40 kg jumps out of a boat of 200 kg on the bank, with a velocity of 2m/ s. If the momentum is conserved. With what velocity the boat will move backwards?​

Answer 1

Explanation: velocity of boat = 40*2/200 = 8/20 =.4

Answer 2

Given:

Boy of mass 40 kg jumps out of a boat of 200 kg on the bank, with a velocity of 2m/s.

To find:

Velocity the boat?

Solution:

Applying Principle of CONSERVATION OF LINEAR MOMENTUM:

$\sf m_{1}u_{1} + m_{2}u_{2} = m_{1}v_{1} + m_{2}v_{2}$

$\sf \implies 0+ 0 =( 40 \times 2) + 200v_{2}$

$\sf \implies 200v_{2} = - 80$

$\sf \implies v_{2} = - 0.4 \: m {s}^{ - 1}$

So, boat moves with 0.4 m/s velocity opposite to direction of boy.

• Negative sign denotes opposite direction.