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Q6.19:
A trolley of mass 300 kg carrying a sandbag of 25 kg is moving uniformly with a speed of 27 km/h on a frictionless track. After a while, sand starts leaking out of a hole on the floor of the trolley at the rate of $0.05 \text{ kg s}^{-1}$. What is the speed of the trolley after the entire sand bag is empty ?

Solution :

Given Variables & Initial Setup

  • Mass of the trolley ($M_t$) = $300\text{ kg}$
  • Initial mass of the sandbag ($M_s$) = $25\text{ kg}$
  • Initial velocity of the trolley-sandbag system ($v_i$) = $27\text{ km/h}$
  • Rate of sand leakage ($\frac{dm}{dt}$) = $0.05\text{ kg s}^{-1}$

First, we perform a unit conversion for velocity into SI units (meters per second):

$v_i = 27 \times \frac{5}{18}\text{ m s}^{-1} = 7.5\text{ m s}^{-1}$ [Conversion factor: $1\text{ km/h} = \frac{1000\text{ m}}{3600\text{ s}}$]

Step 1: Analysis of External Forces

The system comprises the trolley and the sandbag moving on a frictionless track. We analyze the forces acting in the direction of motion:

  • The track is frictionless, implying there is no external horizontal force ($F_{ext} = 0$).
  • The sand drops vertically through the floor of the trolley. Since the sand has no initial horizontal velocity relative to the trolley as it exits, it carries away no momentum in the horizontal direction.

[Per Newton's Second Law in the form $F_{ext} = \frac{dp}{dt}$, if $F_{ext} = 0$, then the linear momentum $p$ of the system must remain constant.]

Step 2: Conservation of Linear Momentum

The momentum of the system at any time $t$ is given by $p = m(t) \cdot v(t)$, where $m(t)$ is the instantaneous mass of the system and $v(t)$ is its velocity. Since the external force is zero, the momentum remains constant throughout the process of the sand leaking.

$P_{initial} = P_{final}$

$(M_t + M_s) \cdot v_i = (M_t + M_{s, final}) \cdot v_f$

Given that the entire sandbag leaks out, the final mass of the sandbag $M_{s, final} = 0$.

Substituting the values:

$(300 + 25) \cdot 7.5 = (300 + 0) \cdot v_f$

Step 3: Calculating Final Velocity

Rearranging the equation to solve for $v_f$:

$v_f = \frac{(300 + 25) \times 7.5}{300}$

$v_f = \frac{325 \times 7.5}{300}$

$v_f = \frac{325}{40}$ [Simplifying $7.5/300 = 1/40$]

$v_f = 8.125\text{ m s}^{-1}$

Vector Analysis and Visual Representation

v = 8.125 m/s Trolley (300 kg)

Note: The rate of leakage ($\frac{dm}{dt}$) is irrelevant to the final velocity calculation because the horizontal velocity of the sand is zero at the moment of separation, resulting in no change to the system's total horizontal momentum.

Final Solution: The speed of the trolley after the entire sandbag is empty remains 7.5 m/s (or 27 km/h), as the sand leaks vertically and exerts no horizontal impulse on the trolley.


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