Question:
A spaceship in space sweeps stationary interplanetary dust. As a result, its mass increases at a rate \(\frac{dM(t)}{dt} = bv^2(t)\), where v(t) is its instantaneous velocity. The instantaneous acceleration of the satellite is :
A spaceship in space sweeps stationary interplanetary dust. As a result, its mass increases at a rate \(\frac{dM(t)}{dt} = bv^2(t)\), where v(t) is its instantaneous velocity. The instantaneous acceleration of the satellite is :
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No external force acts, so the momentum of the (ship + swept dust) is conserved: \(\frac{d(Mv)}{dt} = 0\).
Updated On: Nov 5, 2026
- \(-bv^3(t)\)
- \(-\frac{2bv^3}{M(t)}\)
- \(-\frac{bv^3}{M(t)}\)
- \(-\frac{bv^3}{2M(t)}\)
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The Correct Option is C
Solution and Explanation
No external force acts on the system, and the dust is initially at rest, so \(\frac{d(Mv)}{dt} = 0\).
\(M\frac{dv}{dt} + v\frac{dM}{dt} = 0\).
Substituting \(\frac{dM}{dt} = bv^2\): \(M a + v\cdot bv^2 = 0\).
\(a = -\frac{bv^3}{M(t)}\) (a deceleration).
Answer: (3) \(-\frac{bv^3}{M(t)}\)
\(M\frac{dv}{dt} + v\frac{dM}{dt} = 0\).
Substituting \(\frac{dM}{dt} = bv^2\): \(M a + v\cdot bv^2 = 0\).
\(a = -\frac{bv^3}{M(t)}\) (a deceleration).
Answer: (3) \(-\frac{bv^3}{M(t)}\)
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