A body of mass 1 kg falls freely from a height of 100 m on a platform of mass 3 kg which is mounted

Q: Why do we ignore the initial compression of the spring due to the platform's weight?

The initial compression $x_0 = Mg/k$ is extremely small (around $0.024\text{ mm}$) because the spring constant is very large ($1.25 \times 10^6\text{ N/m}$). It's negligible compared to the compression caused by the high-speed impact.

Q: Why is momentum conserved during the collision but not mechanical energy?

The collision is perfectly inelastic (the mass sticks to the platform), meaning kinetic energy is lost as heat and sound. However, because the collision happens almost instantaneously, external forces like gravity and the spring force don't have time to change the system's momentum.

Q: Why do we ignore the change in gravitational potential energy during the spring's compression?

The compression $x$ is very small ($2\text{ cm}$). The loss in gravitational potential energy $(m+M)gx$ is about $0.8\text{ J}$, which is tiny compared to the $250\text{ J}$ of kinetic energy converted into spring energy.

Q: Could we solve this using the work-energy theorem from the very beginning?

No, because the collision is inelastic. You must use momentum conservation for the collision phase to account for the loss of kinetic energy.

Q: What if the collision was perfectly elastic?

If it were perfectly elastic, the mass would bounce back, and the platform would acquire a different velocity. You would use both momentum and kinetic energy conservation for the collision phase.

Question

A body of mass $1	ext{ kg}$ falls freely from a height of $100	ext{ m}$ on a platform of mass $3	ext{ kg}$ which is mounted on a spring having spring constant $k = 1.25 	imes 10^6	ext{ N/m}$. The body sticks to the platform and the spring's maximum compression is found to be $x$. Given that $g = 10	ext{ ms}^{-2}$, the value of $x$ will be close to

Options:

(A) $8	ext{ cm}$

(B) $4	ext{ cm}$

(C) $40	ext{ cm}$

(D) $80	ext{ cm}$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing in a massive testing facility. High above you, at a staggering height of $100 	ext{ m}$, a $1 	ext{ kg}$ block is held perfectly still.

Directly below it, resting on the ground, is a $3 	ext{ kg}$ platform supported by an incredibly stiff industrial spring. The spring constant is a massive $k = 1.25 	imes 10^6 	ext{ N/m}$.

Our mission is to determine exactly how much this heavy-duty spring will compress when the block is dropped and slams into the platform. This problem is a beautiful symphony of kinematics, momentum, and energy conservation. Let's break it down step by step.

### Phase 1: The Free Fall

The moment the block is released, gravity takes over. As it plummets, its gravitational potential energy is rapidly converted into kinetic energy.

We need to find its velocity exactly at the instant before it strikes the platform. Since it starts from rest, we can use the classic kinematic equation for an object in free fall:

$$v = \sqrt{2gh}$$

Let's plug in our known values. The acceleration due to gravity $g$ is $10 	ext{ m/s}^2$, and the height $h$ is $100 	ext{ m}$.

$$v = \sqrt{2 	imes 10 	imes 100} = \sqrt{2000} = 20\sqrt{5} 	ext{ m/s}$$

So, right before impact, the $1 	ext{ kg}$ block is hurtling downwards at $20\sqrt{5} 	ext{ m/s}$.

### Phase 2: The Inelastic Collision

Now comes the violent part—the collision. The block hits the platform and sticks to it. This is a **perfectly inelastic collision**.

Because the impact happens in a fraction of a millisecond, the external forces (like gravity and the spring force) don't have time to significantly alter the system's momentum during the crash. Therefore, we can safely apply the principle of conservation of linear momentum.

The momentum just before the crash must equal the momentum just after the crash:

$$m v = (m + M) V$$

Here, $V$ is the new velocity of the combined mass system. Let's substitute our values:

$$1 	imes 20\sqrt{5} = (1 + 3) V$$

$$20\sqrt{5} = 4 V \implies V = 5\sqrt{5} 	ext{ m/s}$$

Immediately after the collision, the combined $4 	ext{ kg}$ mass begins moving downwards at $5\sqrt{5} 	ext{ m/s}$.

### Phase 3: The Spring Compression

Now for the final act. The combined mass pushes down on the platform, compressing the spring.

During this phase, the kinetic energy of the moving masses is converted entirely into the elastic potential energy of the spring.

*Note: Technically, the mass also loses a tiny bit of gravitational potential energy as it moves down by the compression distance $x$. However, because the spring is so incredibly stiff, $x$ will be very small, making this gravitational energy loss negligible compared to the massive kinetic energy involved.*

We equate the kinetic energy to the spring's potential energy:

$$\frac{1}{2}(m + M)V^2 = \frac{1}{2}kx^2$$

Let's substitute the values into our master equation:

$$\frac{1}{2} (4) (5\sqrt{5})^2 = \frac{1}{2} (1.25 	imes 10^6) x^2$$

Simplifying the left side, the square of $5\sqrt{5}$ is $125$. Multiplying by $2$ gives us the total kinetic energy:

$$250 = \frac{1}{2} (1.25 	imes 10^6) x^2$$

$$250 = 625000 x^2$$

### The Final Calculation

We are now one algebraic step away from the truth. Let's solve for $x^2$:

$$x^2 = \frac{250}{625000} = \frac{1}{2500} = 4 	imes 10^{-4} 	ext{ m}^2$$

Taking the square root of both sides gives us the maximum compression:

$$x = \sqrt{4 	imes 10^{-4}} = 2 	imes 10^{-2} 	ext{ m}$$

Converting this to centimeters, we get:

**$$x = 2 	ext{ cm}$$**

The spring compresses by exactly $2 	ext{ cm}$. Interestingly, if you look at the options provided in the original exam question, none of them are $2 	ext{ cm}$. This was a famous bonus question where all given options were incorrect!

Select Answer:

Visualized Solution

\text{Analyzing the Physical System}

\text{Velocity Before Impact}

\text{Calculating Impact Velocity}

\text{Perfectly Inelastic Collision}

\text{Velocity of Combined Mass}

\text{Compression of the Spring}

\text{Equating Energies}

\text{Maximum Compression}

The Sigma Insight: Conservation of Mechanical Energy

Analyzing the Setup

Phase 1

Phase 2

Phase 3

The Final Calculation

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