A railway wagon (open at the top) of mass M₁ is moving with speed v₁ along a straight track. As a result of rain, after some time it gets partially filled with water so that the mass of the wagon becomes M₂ and speed becomes v₂. Taking the rain to be falling vertically and water stationary inside the wagon, the relation between the two speeds v₁ and v₂ is:
- (a)v₁ = v₂
- (b)½ M₁v₁² < ½ M₂v₂²
- (c)M₁v₁ = M₂v₂
- (d)M₁v₁ < M₂v₂
Correct — C, M₁v₁ = M₂v₂. The rain falls vertically, so it adds mass but no horizontal momentum, and no external horizontal force acts on the wagon-plus-water system. Horizontal linear momentum is therefore conserved: the initial horizontal momentum M₁v₁ equals the final horizontal momentum M₂v₂. Because M₂ is larger, this also means the wagon slows down (v₂ < v₁).
- (a)v₁ = v₂ — Adding mass that carries no forward momentum must slow the wagon to keep momentum constant, so v₂ is less than v₁, not equal.
- (b)½ M₁v₁² < ½ M₂v₂² — Kinetic energy actually decreases here. Since momentum p is conserved, KE = p²/2M and M increases, so ½M₂v₂² is less than ½M₁v₁² — the inequality is the wrong way round (energy is lost in the inelastic capture of water).
- (d)M₁v₁ < M₂v₂ — Horizontal momentum is conserved, not increased; there is no forward force to make the final momentum exceed the initial. So it is an equality, M₁v₁ = M₂v₂, not a 'less than'.
This is a variable-mass (inelastic accretion) problem governed by conservation of linear momentum. When mass is added with zero velocity in the direction of motion and no external force acts along that direction, the momentum in that direction stays constant while the speed drops. Kinetic energy is not conserved because capturing the water is an inelastic process.
The key is to separate the vertical and horizontal directions. Vertically, gravity and the rain act, but horizontally the rain adds no momentum and nothing pushes the wagon. So only horizontal momentum is conserved, giving M₁v₁ = M₂v₂ — and, as a check, the loss of kinetic energy rules out option (b).
- Rain falling vertically adds mass but no horizontal momentum.
- With no horizontal external force, horizontal momentum is conserved: M₁v₁ = M₂v₂.
- Hence v₂ = (M₁/M₂)·v₁, which is less than v₁ since M₂ > M₁.
- Kinetic energy falls (KE = p²/2M with p fixed and M rising) — the accretion is inelastic.
Only horizontal momentum is conserved, giving M₁v₁ = M₂v₂ — option (c).
- Assuming speed is unchanged when mass is added — momentum, not speed, is conserved.
- Thinking kinetic energy is conserved; the inelastic capture of water loses energy, so option (b) is reversed.
Asked as a conceptual mechanics item on which quantity is conserved when a moving body gains stationary mass.
No directly related past PYQ was found.
- practice — not a real PYQ
A moving trolley gains mass by collecting sand dropped vertically onto it (no horizontal force). Its speed:
- (a)increases
- (b)decreases
- (c)stays the same
- (d)becomes zero
Answer(b) decreases — horizontal momentum is conserved while mass rises.
- practice — not a real PYQ
When a wagon collects vertically falling rain, which quantity of the wagon-water system stays constant horizontally?
- (a)kinetic energy
- (b)speed
- (c)linear momentum
- (d)acceleration
Answer(c) linear momentum — since no horizontal external force acts.