A cylindrical solid of mass 10^-2 kg and cross-sectional area 10^-4 m^2 is moving parallel to its ax

Q: Wait, why is there no external force? What about the force required to accelerate the dust?

The force required to accelerate the dust is an *internal* force between the cylinder and the dust particles. If we take the cylinder and the dust cloud as our system, there is no net external force acting on them in the horizontal direction.

Q: Why can't we just use $F = ma$ here?

Newton's second law in the form $F = ma$ is only valid for systems with constant mass. For a variable mass system, you must use the more general form $F = \frac{dp}{dt}$. Since $F_{ext} = 0$, momentum $p$ is constant.

Q: Why do we integrate from $0$ to $x$ and $0$ to $t$?

The problem states that at $t = 0$, the front face of the cylinder is at $x = 0$. These are our initial boundary conditions. We want to find the position $x$ at a later time $t$, so we integrate from the initial state to the final state.

Q: Is kinetic energy conserved in this process?

No! This is a perfectly inelastic collision process. The dust particles stick to the cylinder, which means a significant amount of kinetic energy is lost as heat. Only momentum is conserved.

Q: Why do we reject the negative root of the quadratic equation?

The cylinder is moving in the positive x-direction with an initial positive velocity. Since there are no forces to turn it around, its position $x$ must be positive for all $t > 0$. The negative root corresponds to a mathematical solution that doesn't fit our physical reality.

Question

A cylindrical solid of mass $10^{-2} 	ext{ kg}$ and cross-sectional area $10^{-4} 	ext{ m}^2$ is moving parallel to its axis (the x-axis) with a uniform speed of $10^3 	ext{ m/s}$ in the positive direction. At $t = 0$, its front face passes the plane $x = 0$. The region to the right of this plane is filled with stationary dust particles of uniform density $10^{-3} 	ext{ kg/m}^3$. When a dust particle collides with the face of the cylinder, it sticks to its surface. Assuming that the dimensions of the cylinder remain practically unchanged and that the dust sticks only to the front face of the cylinder find the x-coordinate of the front of the cylinder at $t = 150 	ext{ s}$.

Sigma Percentile · Accepted Answer

The physics of variable mass systems is one of the most fascinating topics in classical mechanics. When we think of Newton's laws, we often picture rigid blocks sliding down inclines or billiard balls colliding. In those cases, the mass of the object remains constant. But what happens when an object gains or loses mass as it moves? Think of a rocket burning fuel, a raindrop accumulating moisture as it falls, or, in our case, a cylinder sweeping through a cosmic dust cloud.

### Analyzing the Setup

Imagine a massive cylindrical solid hurtling through space along the x-axis. At exactly $t = 0$, its front face crosses the boundary $x = 0$ and enters a region filled with stationary dust particles. As the cylinder moves forward, these dust particles collide with its front face and stick to it.

This means the mass of our cylinder is not constant; it is continuously increasing! However, there is a crucial physical constraint here: there are no external forces acting on the cylinder-dust system in the horizontal direction. The force of the dust hitting the cylinder is an internal force. Because the net external force is zero, the total linear momentum of the system must be conserved.

### The Master Equation

Let's define the mass of the cylinder at any position $x$. The initial mass is $m_0$. As the cylinder moves a distance $x$, it sweeps out a volume equal to its cross-sectional area $A$ multiplied by $x$. The mass of the dust in this volume is the density $ho$ times the volume. Therefore, the mass of the cylinder as a function of position $x$ is:

$$m(x) = m_0 + ho A x$$

Now, we apply the principle of conservation of linear momentum. The initial momentum of the cylinder before it hits the dust is $m_0 v_0$. At some later position $x$, its mass is $m(x)$ and its velocity is $v$. Equating the initial and final momentum gives us:

$$m_0 v_0 = (m_0 + ho A x) v$$

We know that velocity $v$ is the rate of change of position, so we can substitute $v = \frac{dx}{dt}$:

$$m_0 v_0 = (m_0 + ho A x) \frac{dx}{dt}$$

To solve this differential equation, we separate the variables $x$ and $t$:

$$(m_0 + ho A x) dx = m_0 v_0 dt$$

Now, we integrate both sides. The position goes from $0$ to $x$, and the time goes from $0$ to $t$:

$$\int_0^x (m_0 + ho A x) dx = \int_0^t m_0 v_0 dt$$

Performing the integration, we get our master equation that relates position and time:

$$m_0 x + \frac{ho A x^2}{2} = m_0 v_0 t$$

### Final Calculation

With our master equation ready, it is time to plug in the given numerical values. We are given $m_0 = 10^{-2} 	ext{ kg}$, $A = 10^{-4} 	ext{ m}^2$, $ho = 10^{-3} 	ext{ kg/m}^3$, $v_0 = 10^3 	ext{ m/s}$, and $t = 150 	ext{ s}$. Substituting these into the equation:

$$10^{-2} x + \frac{(10^{-3})(10^{-4}) x^2}{2} = (10^{-2})(10^3)(150)$$

Simplifying the terms, we get:

$$10^{-2} x + \frac{10^{-7}}{2} x^2 = 1500$$

To make this equation easier to solve, we can multiply the entire equation by $2 	imes 10^7$ to clear the fractions and negative exponents:

$$2 	imes 10^5 x + x^2 = 3000 	imes 10^7$$

Rearranging this into a standard quadratic equation form:

$$x^2 + 2 	imes 10^5 x - 3 	imes 10^{10} = 0$$

We can solve this quadratic equation using the standard quadratic formula.

$$x = \frac{-2 	imes 10^5 \pm \sqrt{(2 	imes 10^5)^2 - 4(1)(-3 	imes 10^{10})}}{2}$$

$$x = \frac{-2 	imes 10^5 \pm \sqrt{4 	imes 10^{10} + 12 	imes 10^{10}}}{2}$$

$$x = \frac{-2 	imes 10^5 \pm \sqrt{16 	imes 10^{10}}}{2}$$

$$x = \frac{-2 	imes 10^5 \pm 4 	imes 10^5}{2}$$

This gives us two possible roots. Since the cylinder is moving in the positive x-direction, its position $x$ must be positive. Therefore, we reject the negative root and take the positive one:

$$x = \frac{2 	imes 10^5}{2} = 10^5 	ext{ m}$$

**The final position of the cylinder's front face is $10^5 	ext{ m}$.**

Visualized Solution

Visualizing the Setup

The Physics of Variable Mass

Mass as a Function of Position

Applying Momentum Conservation

Introducing Calculus

Separation of Variables

Setting Up the Integral

Performing the Integration

Substituting the Given Values

Simplifying the Equation

Forming the Quadratic Equation

Solving for Position

The Sigma Insight: Conservation of Linear Momentum

Analyzing the Setup

The Master Equation

Final Calculation

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