A trolley of mass $300\ \text{kg}$ carrying a sandbag of $25\ \text{kg}$ is moving uniformly with a speed of $27\ \text{km/h}$ 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?

A trolley of mass 300kg carrying a sandbag of 25kg is moving uniformly with a speed of 27km/h on a frictionles

Exercises · 5.19

300\ \text{kg}

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Concept understanding — Conservation of Momentum

Conservation of Momentum: From Push to Principle

Imagine you're standing on perfectly smooth ice, wearing skates. You're completely still. Now, you push a heavy medicine ball away from you. What happens? You roll backward. The harder you push the ball, the faster you roll back.

That's the core intuition: you can't push something away without being pushed back yourself. The push you give the ball is matched by an equal push on you, in the opposite direction. This isn't a special property of ice or skates — it's a fundamental rule of how forces work in the universe.

The Hidden Quantity That Never Changes

Physicists call the "amount of motion" an object has its momentum. For everyday speeds, momentum is simple:

\vec{p} = m \vec{v}

Where $m$ is mass (how much stuff) and $\vec{v}$ is velocity (speed with direction). Momentum is a vector — it cares about which way you're going.

A truck creeping forward has huge momentum (big mass, small speed). A bullet zipping through air has moderate momentum (tiny mass, huge speed). A parked car has zero momentum (speed is zero).

Now here's the key: in any isolated system (no outside forces), total momentum stays the same. Always. Before, during, and after any interaction.

The Precise Statement

Important

Law of Conservation of Momentum:

In a closed, isolated system (no external forces), the total vector momentum of the system remains constant over time.

Mathematically, for two objects that interact (collide, push apart, explode):

\vec{p}_{1,\text{initial}} + \vec{p}_{2,\text{initial}} = \vec{p}_{1,\text{final}} + \vec{p}_{2,\text{final}}

Or in terms of masses and velocities:

m_1 \vec{u}_1 + m_2 \vec{u}_2 = m_1 \vec{v}_1 + m_2 \vec{v}_2

Where $\vec{u}$ means initial velocity and $\vec{v}$ means final velocity.

Why This Works: Newton's Third Law in Disguise

When you push the medicine ball, your hand exerts a force $\vec{F}$ on the ball. By Newton's Third Law, the ball exerts an equal and opposite force $-\vec{F}$ back on your hand. These forces act for the same time $\Delta t$ .

Force times time equals impulse, which equals change in momentum:

\vec{F} \Delta t = \Delta \vec{p}

For you and the ball:

Ball's momentum change: $+\vec{F} \Delta t$ (ball goes forward)
Your momentum change: $-\vec{F} \Delta t$ (you go backward)

Add them: $+\vec{F} \Delta t + (-\vec{F} \Delta t) = 0$

Total change is zero. Momentum is conserved because forces always come in equal-and-opposite pairs.

Note

This is why a rocket works in the vacuum of space. It throws exhaust backward (one momentum change), and the rocket itself moves forward (equal opposite momentum change). No air needed — just Newton's Third Law and conservation of momentum.

What This Law Does NOT Mean

It does NOT mean individual objects keep constant momentum. Only the total of all objects in the system stays constant. Individual momenta can change wildly.
It does NOT apply if external forces act. If friction, gravity from outside, or a wall stops something, momentum is not conserved for that system. (You can expand the system to include the Earth or the wall, and then momentum is conserved again.)
It does NOT require collisions to be elastic. Even in a messy, sticky, energy-losing collision, momentum is still perfectly conserved. Energy can be lost to heat or deformation, but momentum never disappears.

A Quick Example …

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The key idea is that on a frictionless track, no external horizontal force acts on the system, so the total horizontal momentum is conserved. The speed of the trolley remains constant at $27\ \text{km/h}$ even as sand leaks out, because the sand leaves with the same horizontal velocity as the trolley. The final answer is $27\ \text{km/h}$ .

This is a classic problem that often trips students up because they instinctively think that losing mass should change the speed — like a rocket ejecting fuel. But the physics here is fundamentally different. Let's see why.

The Concept: Conservation of Momentum

Momentum is conserved in a system when no external force acts on it. Here, the track is frictionless, so there is no horizontal force from outside. The leaking sand is part of the system — it doesn't push against the trolley as it leaves. Instead, it simply drops out vertically (relative to the trolley) and continues moving forward with the same horizontal speed it had just before leaving.

Watch out

A common mistake is to treat this like a rocket, where ejected mass is pushed backward, creating thrust. Here, sand leaks out passively — it doesn't get thrown. So there's no recoil. The trolley's speed does not increase.

Let's work through it step by step.

Identify the system and initial conditions The system is the trolley + the sandbag (all the sand inside it). Total initial mass:

m_i = 300\ \text{kg} + 25\ \text{kg} = 325\ \text{kg}

Initial speed:

v_i = 27\ \text{km/h}

Since the track is frictionless, no external horizontal force acts. So total horizontal momentum $P$ is constant.

What happens as sand leaks?

Sand falls out through a hole in the floor. At the instant a grain of sand leaves, it still has the same forward velocity as the trolley (because it was moving with it). It doesn't get pushed backward or forward relative to the trolley — it just drops straight down relative to the trolley's frame.

So the sand that leaves carries away momentum equal to $(\text{mass of sand}) \times v_{\text{trolley at that instant}}$ .
Apply conservation of momentum

Let $M(t)$ be the mass of the trolley + remaining sand at time $t$ , and $v(t)$ be its speed. The momentum of the system at any time is:

P(t) = M(t)\,v(t) + \int (\text{momentum of sand that has left})

But here's the neat part: because the sand leaves with exactly the same horizontal velocity as the trolley, the momentum carried away is exactly the same as if that sand had stayed on board. So the total momentum of the system (trolley + all sand, whether on board or already fallen) remains constant and equal to the initial momentum. …

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The Concept: Conservation of Momentum

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