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Question

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 the water stationery inside the wagon, the relation between the two speeds v1 and v2 is :

This question was previously asked in
NDA I 2023 GAT Previous Year Paper (16-Apr-2023)
The correct answer is M1v 1 = M2v 2

Understanding Momentum Conservation in a Railway Wagon Problem

This problem involves analyzing the motion of a railway wagon as its mass changes due to falling rain. The key physical principle at play here is the conservation of momentum. Let's break down the scenario.

Analyzing the Scenario and Forces

We have a railway wagon moving along a straight track. Rain is falling vertically into the wagon. The rain adds mass to the wagon, increasing its total mass and potentially changing its speed.

  • Initial state: Wagon mass \({M_1}\), speed \({v_1}\).
  • Final state: Wagon mass \({M_2}\) (where \({M_2 > M_1}\) because of the added water), speed \({v_2}\).

The rain is falling vertically. This means the force exerted by the rain on the wagon as it lands is primarily vertical. While there might be a vertical impulse changing the vertical momentum (causing the wagon to settle slightly on its springs, if any), there is no external force acting on the wagon-water system in the horizontal direction.

Since there is no external horizontal force, the total horizontal momentum of the system (wagon + collected rain) must be conserved.

Applying Conservation of Horizontal Momentum

Momentum (\({p}\)) is defined as the product of mass (\({m}\)) and velocity (\({v}\)), i.e., \({p = mv}\). Since we are considering only the horizontal motion, we will look at the horizontal components of momentum.

  • Initial horizontal momentum of the wagon: \({p_{1x} = M_1 v_1}\). The rain falling vertically has zero initial horizontal momentum before it lands in the wagon.
  • Final horizontal momentum of the wagon plus collected water: As the rain lands and becomes part of the wagon's mass, the combined system (wagon + water) moves with the final speed \({v_2}\). So, the final horizontal momentum is \({p_{2x} = M_2 v_2}\).

According to the principle of conservation of linear momentum in the absence of external horizontal forces:

Initial horizontal momentum = Final horizontal momentum

\({p_{1x} = p_{2x}}\)

\({M_1 v_1 = M_2 v_2}\)

Comparing with the Given Options

Let's look at the provided options and compare them with our derived relation:

  1. \({v_1 = v_2}\): This would only be true if \({M_1 = M_2}\). Since mass increases due to rain (\({M_2 > M_1}\)), this option is incorrect.
  2. \({&frac12;M_1 v_1^2 < &frac12; M_2 v_2^2}\): This relates to kinetic energy. In this scenario, kinetic energy is not conserved. As the rain with zero initial horizontal kinetic energy is accelerated horizontally to the wagon's speed, some kinetic energy is lost in the inelastic collision of the rain with the wagon and the subsequent internal thermalization within the water. The final kinetic energy should be less than the initial kinetic energy, making this option incorrect.
  3. \({M_1v_1 = M_2v_2}\): This matches our derived relation based on the conservation of horizontal momentum.
  4. \({M_1v_1 < M_2v_2}\): This implies that the final horizontal momentum is greater than the initial horizontal momentum. This would require an external horizontal force doing positive work, which is not present in this problem. Thus, this option is incorrect.

Therefore, the correct relation between the initial and final speeds and masses is \({M_1v_1 = M_2v_2}\).

Conclusion

Based on the principle of conservation of horizontal momentum, as there is no external horizontal force acting on the wagon-water system, the total horizontal momentum remains constant. The initial horizontal momentum is \({M_1v_1}\) and the final horizontal momentum is \({M_2v_2}\). Equating these gives the relation \({M_1v_1 = M_2v_2}\).

Revision Table: Railway Wagon Momentum

Concept Initial State Final State Relation
Mass \({M_1}\) \({M_2}\) (\({M_2 > M_1}\)) Mass increases
Speed \({v_1}\) \({v_2}\) Speed changes
Horizontal Momentum \({M_1 v_1}\) \({M_2 v_2}\) Conserved (\({M_1 v_1 = M_2 v_2}\))
External Horizontal Force None None Zero

Additional Information: Conservation of Momentum

Conservation of linear momentum is a fundamental principle in physics. It states that if no external forces act on a system, the total linear momentum of the system remains constant.

  • Linear Momentum: Defined as the product of mass and velocity (\({p = mv}\)). It is a vector quantity, having both magnitude and direction.
  • External vs. Internal Forces: External forces are forces exerted on the system by objects outside the system. Internal forces are forces exerted by objects within the system on each other. Conservation of momentum applies when the net external force is zero.
  • Application: This principle is widely used to analyze collisions and explosions, rocket propulsion, and systems where mass changes over time, like the wagon filling with rain or a leaking tank.
  • Change in Kinetic Energy: While momentum is conserved in the absence of external forces, kinetic energy is not necessarily conserved. In inelastic processes like the rain collecting in the wagon (where kinetic energy is transformed into other forms like heat and sound), the total kinetic energy of the system changes.
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