Two particles of masses m1 and m2 in projectile motion have velocities v1 and v2 respectively at tim

Q: Wait, why don't we need to know the type of collision (elastic or inelastic)?

Because the forces exerted by the particles on each other during the collision are internal to the two-particle system. By Newton's Third Law, these forces are equal and opposite, so they cancel out perfectly when calculating the *total* momentum of the system. The nature of the collision only affects individual velocities, not the system's total momentum.

Q: Why is gravity the only external force? What about the force of the collision?

The collision force is an *internal* force because it acts between the two particles that make up our defined system. An external force must come from outside the system. Since the particles are in mid-air, the only external entity exerting a force on them is the Earth (gravity).

Q: Can we just use the equations of kinematics ($v = u + at$) here?

Absolutely! In fact, the Impulse-Momentum theorem for a constant force is mathematically identical to the first equation of motion. If you write $\mathbf{v}_f = \mathbf{v}_i + \mathbf{g}t$ for both particles and add their momenta, you will arrive at the exact same result. The impulse method is just a more elegant way to view it.

Q: Why is the time interval $2t_0$ and not just $t_0$?

The question asks for the difference between the momentum at time $t = 2t_0$ (when velocities are $\mathbf{v}_1'$ and $\mathbf{v}_2'$) and the momentum at time $t = 0$ (when velocities are $\mathbf{v}_1$ and $\mathbf{v}_2$). Therefore, the total elapsed time is $2t_0 - 0 = 2t_0$.

Q: What if the particles had different masses, say $m_1$ and $10m_2$? Would the logic change?

Not at all! The beauty of this principle is that it is entirely independent of the individual masses or their ratio. The net external force would simply be $(m_1 + 10m_2)\mathbf{g}$, and the logic would flow exactly the same way.

Question

Two particles of masses $m_1$ and $m_2$ in projectile motion have velocities $\mathbf{v}_1$ and $\mathbf{v}_2$ respectively at time $t = 0$. They collide at time $t_0$. Their velocities become $\mathbf{v}_1'$ and $\mathbf{v}_2'$ at time $2t_0$ while still moving in air. The value of $|(m_1 \mathbf{v}_1' + m_2 \mathbf{v}_2') - (m_1 \mathbf{v}_1 + m_2 \mathbf{v}_2)|$ is

Options:

(A) zero

(B) $(m_1 + m_2)g t_0$

(C) $2 (m_1 + m_2)g t_0$

(D) $\frac{1}{2}(m_1 + m_2)g t_0$

Sigma Percentile · Accepted Answer

### The Illusion of Complexity

When you first look at this problem, it seems like a nightmare of vector algebra. The expression $|(m_1 \mathbf{v}_1' + m_2 \mathbf{v}_2') - (m_1 \mathbf{v}_1 + m_2 \mathbf{v}_2)|$ looks like it requires you to track the individual trajectories, calculate the exact collision dynamics, and then perform a messy vector subtraction.

But physics is often about seeing through the illusion of complexity. Let's break down what this expression actually means. The term $(m_1 \mathbf{v}_1' + m_2 \mathbf{v}_2')$ is simply the total final momentum of the two-particle system at time $t = 2t_0$. Similarly, $(m_1 \mathbf{v}_1 + m_2 \mathbf{v}_2)$ is the total initial momentum at $t = 0$.

Therefore, the entire expression is just asking for the magnitude of the change in the total momentum of the system, $|\Delta \mathbf{P}|$. Once we realize this, the problem transforms from a tedious calculation into a beautiful conceptual puzzle.

### The Power of the System

To solve this elegantly, we must define our system as the collection of both particles, $m_1$ and $m_2$. Why is this so powerful? Because it allows us to completely ignore the collision!

When the particles collide at time $t = t_0$, they exert massive forces on each other. However, according to Newton's Third Law, these forces are equal and opposite. Because both particles are inside our defined system, these collision forces are **internal forces**.

Internal forces can violently alter the individual velocities of the particles, but they can never, ever change the total momentum of the system. They perfectly cancel each other out.

### The Master Equation

If internal forces can't change the system's momentum, what can? Only **external forces**.

As the particles fly through the air, before, during, and after the collision, there is only one external force acting on them: the gravitational pull of the Earth.

According to Newton's Second Law for a system of particles, the rate of change of total momentum is equal to the net external force:

$$ \frac{d\mathbf{P}}{dt} = \mathbf{F}_{ext} $$

The net external force is simply the sum of the weights of the two particles:

$$ \mathbf{F}_{ext} = m_1\mathbf{g} + m_2\mathbf{g} = (m_1 + m_2)\mathbf{g} $$

Notice that this force is completely constant. It doesn't care about the collision; it just keeps pulling downwards with the exact same strength.

### The Final Calculation

To find the total change in momentum, we use the Impulse-Momentum theorem, which is just the integral of Newton's Second Law over time:

$$ \Delta \mathbf{P} = \int \mathbf{F}_{ext} dt $$

Because our external force is constant, we can pull it out of the integral, simplifying the equation to:

$$ \Delta \mathbf{P} = \mathbf{F}_{ext} \Delta t $$

Now, we must be careful with the time interval. The problem asks for the difference between the momentum at $t = 2t_0$ and $t = 0$. Therefore, the total elapsed time is $\Delta t = 2t_0 - 0 = 2t_0$.

Substituting our values into the impulse equation, we get:

$$ \Delta \mathbf{P} = (m_1 + m_2)\mathbf{g} (2t_0) $$

The question asks for the magnitude of this change. Since the magnitude of the gravity vector $\mathbf{g}$ is simply $g$, we arrive at our final, elegant answer:

$$ |\Delta \mathbf{P}| = 2(m_1 + m_2)g t_0 $$

By zooming out and looking at the system as a whole, we bypassed all the messy collision math and solved the problem in just a few lines of logic!

Select Answer:

Visualized Solution

Decoding the Expression

Internal vs External Forces

The Role of Gravity

Newton's Second Law for a System

Net External Force

Impulse-Momentum Theorem

The Time Interval

Calculating the Magnitude

The Sigma Insight: Conservation of Linear Momentum

The Illusion of Complexity

The Power of the System

The Master Equation

The Final Calculation

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