JEE Main 2020 · 5 September · evening · Single correct · +4 / −1
A spaceship in space sweeps stationary interplanetary dust. As a result, its mass
increases at a rate dM( t )/dt = bv²(t), where v(t) is its instantaneous velocity. The instantaneous acceleration of the satellite is :
- (a) -bv³(t)
- (b) - 2bv³/M( t )
- (c) - bv³/M( t )
- (d) - bv³/2M( t )
Answer and solution
Answer (c) - bv³/M( t )
Solution
Physics sir · 3 steps
No external force acts, so the momentum Mv stays constant. Differentiate it and the acceleration follows.
- Momentum is conserved. Take the ship and the dust it sweeps up as one system. The dust starts at rest, so it brings in no momentum, and no external force acts in space. So the total momentum M(t) v(t) stays constant.
d/dt(Mv) = Mdv/dt + vdM/dt = 0
- Put in the rate. Substitute dM/dt = bv². This gives Mdv/dt = -v × bv² = -bv³.
- So. The acceleration is dv/dt = -bv³/M(t), which is option (c). The minus sign means the ship slows down as it picks up dust.
The idea to keep. When a body gains mass from material at rest and no external force acts, write Mv = constant and differentiate. The mass gain shows up as a drag, M dv/dt = -v dM/dt.
Answer from the NTA final answer key. Solution by PrepWiser.