Question

Q23. A 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? O (a) 0.4 m/s O (b) 0.8 m/s O (c) 5 m/s о (d) 10 m/s​

Answer 1

Answer:

option a) 0.4 m/s is the correct answer

Explanation:

Velocity of the Boat=

$\frac{40 \times 2}{200}$

$\frac{80}{200} \\ = \frac{40}{100} \\ = 0.4$

hope it helps you

Answer 2

Answer:

0.4 m/s

explanation:

if u = velocity of the boat

40 kg * 2 = 200 kg * u [ third law of motion ]

80 = 200kg * u

80/200 = u

0.4 m/s = u