A man weighing 50 kg jumps off a stationary boat of mass 250 kg into water with a horizontal velocity 2 m s⁻¹ relative to the ground. Assuming no friction with water, the recoil speed of the boat is:
A0.2 m s⁻¹
B0.4 m s⁻¹
C0.8 m s⁻¹
D1.0 m s⁻¹
Answer & Solution
Correct answer: B. 0.4 m s⁻¹
1. The man + boat system starts at rest, so total momentum = 0.
2. After the jump: m_man v_man + m_boat v_boat = 0.
3. 50 × 2 + 250 × v_boat = 0 ⇒ v_boat = −100 / 250 = −0.4 m s⁻¹.
4. The negative sign means the boat recoils opposite to the man's direction.
5. Magnitude = 0.4 m s⁻¹.
_Source: NCERT Class 11 Physics Ch 4 "Laws of Motion", §4.7 Conservation of Momentum_
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