A particle of mass m is dropped from a height h above the ground. At the same time another particle

Q: Why did we use relative motion instead of solving their position equations?

While solving position equations works perfectly fine, relative motion simplifies the math significantly. Since both particles experience the same gravitational acceleration, their relative acceleration is zero, turning a quadratic problem into a simple linear one.

Q: Why does the combined mass come to rest after the collision?

Because the collision is perfectly inelastic, they stick together. We found that their velocities just before impact were equal and opposite. Since their masses are also equal, their total initial momentum was zero. To conserve momentum, the final velocity must also be zero.

Q: Does the fact that the combined mass is '2m' affect its time to fall?

No! In free fall, the acceleration due to gravity 'g' is independent of the object's mass. A 2m mass falls at the exact same rate as an m mass.

Q: What would happen if the collision was perfectly elastic instead?

If the collision were perfectly elastic, they would exchange velocities. The top particle would bounce back up, and the bottom particle would be driven downwards. They wouldn't stick together.

Q: How did we determine the height H where the collision occurred?

We calculated the distance the top particle fell during the time 't_0' using the equation h' = 1/2 g t_0^2. Subtracting this from the total height 'h' gave us the height from the ground, H.

Question

A particle of mass $m$ is dropped from a height $h$ above the ground. At the same time another particle of the same mass is thrown vertically upwards from the ground with a speed of $\sqrt{2gh}$. If they collide head-on completely inelastically, then the time taken for the combined mass to reach the ground, in units of $\sqrt{\frac{h}{g}}$ is

Options:

(A) $\sqrt{\frac{1}{2}}$

(B) $\frac{1}{2}$

(C) $\sqrt{\frac{3}{2}}$

(D) $\sqrt{\frac{3}{4}}$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing on the ground, watching two identical particles. One is dropped from a height $h$, starting its journey with zero initial velocity. At the exact same moment, the second particle is launched upwards from the ground with a formidable speed of $\sqrt{2gh}$.

These two particles are on a collision course. Because they are moving along the same vertical line, they are bound to meet. Our first goal is to figure out *when* and *where* this dramatic encounter takes place.

### The Magic of Relative Motion

We could solve for their positions individually and equate them, but there is a much more elegant way: **Relative Motion**.

Let's observe the top particle from the perspective of the bottom particle. Since both particles are in free fall, they both experience the same downward acceleration due to gravity, $g$. Therefore, their relative acceleration is zero!

$$a_{	ext{rel}} = g - g = 0$$

This means that from the perspective of one particle, the other is approaching at a constant speed. The relative velocity is simply the initial speed of the bottom particle, because the top one started from rest.

$$v_{	ext{rel}} = \sqrt{2gh} - 0 = \sqrt{2gh}$$

Since the relative velocity is constant, the time taken to cover the initial separation distance $h$ is just distance divided by speed.

$$t_0 = \frac{h}{v_{	ext{rel}}} = \frac{h}{\sqrt{2gh}} = \sqrt{\frac{h}{2g}}$$

### The Moment of Impact

Now that we know *when* they collide, let's find out how fast they are going right before the crash. We can use the first equation of motion, $v = u + at$, for each particle.

For the top particle (falling down):
$$v_A = gt_0 = g\sqrt{\frac{h}{2g}} = \sqrt{\frac{gh}{2}}$$

For the bottom particle (moving up):
$$v_B = u - gt_0 = \sqrt{2gh} - g\sqrt{\frac{h}{2g}} = \sqrt{2gh} - \sqrt{\frac{gh}{2}} = \sqrt{\frac{gh}{2}}$$

Look at that! Both particles have the exact same speed just before they collide. This symmetry is a beautiful consequence of the initial conditions.

### The Inelastic Collision

The problem states that the collision is **completely inelastic**. This means the two particles stick together and move as a single combined mass of $2m$.

Let's apply the principle of conservation of linear momentum. Just before the collision, the top particle has momentum $m\sqrt{\frac{gh}{2}}$ downwards, and the bottom particle has momentum $m\sqrt{\frac{gh}{2}}$ upwards.

$$p_{	ext{initial}} = m\sqrt{\frac{gh}{2}} - m\sqrt{\frac{gh}{2}} = 0$$

Since the total initial momentum is zero, the final momentum must also be zero. Therefore, the combined mass momentarily comes to a complete halt right after the collision!

$$v_{	ext{final}} = 0$$

### The Final Fall

To find out how long it takes for this new combined mass to reach the ground, we need to know its height. Let's calculate how far the top particle fell before the collision.

$$h' = \frac{1}{2}gt_0^2 = \frac{1}{2}g\left(\sqrt{\frac{h}{2g}}ight)^2 = \frac{h}{4}$$

So, the collision happens at a distance of $h/4$ from the top. The height from the ground is:

$$H = h - \frac{h}{4} = \frac{3h}{4}$$

Now, the combined mass of $2m$ falls freely from rest from this height $H$. We use the second equation of motion again to find the time of fall, $t$.

$$H = \frac{1}{2}gt^2$$

$$\frac{3h}{4} = \frac{1}{2}gt^2$$

$$t^2 = \frac{3h}{2g}$$

$$t = \sqrt{\frac{3}{2}}\sqrt{\frac{h}{g}}$$

And there we have it! The total time taken for the combined mass to reach the ground is $\sqrt{\frac{3}{2}}$ units of $\sqrt{\frac{h}{g}}$.

Select Answer:

Visualized Solution

$u_{A} = 0, u_{B} = 2 g h$

$v_{rel} = 2 g h$

$t_{0} = \frac{h}{2 g}$

$v_{A} = v_{B} = \frac{g h}{2}$

$v_{final} = 0$

$H = \frac{3 h}{4}$

$t = \frac{3 h}{2 g}$

$t = \frac{3}{2} \frac{h}{g}$

The Sigma Insight: Head-on Collision

Analyzing the Setup

The Magic of Relative Motion

The Moment of Impact

The Inelastic Collision

The Final Fall

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Select Answer:

Visualized Solution

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