A ball of mass 4 kg, moving with a velocity of 10 ms^-1, collides with a spring of length 8 m and fo

Q: Why on earth do we just assume the ball stops completely? Doesn't it bounce back?

It absolutely does bounce back! But we are analyzing the exact instant of *maximum compression*. At that specific microsecond, the ball has to stop moving forward before it can reverse its direction. At that momentary stop, its velocity is zero, meaning all its kinetic energy has been temporarily stored in the spring.

Q: Why can't I just use my trusty kinematics equations like $v^2 = u^2 + 2as$ to find the distance?

Kinematics equations only work when acceleration is constant. Here, the force exerted by the spring is $F = -kx$. Because the force changes depending on how much the spring is compressed ($x$), the acceleration is constantly changing. Therefore, standard kinematics fail, and we must use the work-energy theorem.

Q: What would happen if the floor wasn't perfectly smooth and had friction?

If friction were present, it would do negative work on the ball, draining energy from the system as heat. The equation would change to $K_i - W_{\text{friction}} = U_f$. Because energy is lost, the spring wouldn't compress as much as it does in our frictionless ideal scenario.

Q: I found the compression is 2, so why is the answer 6? This is so confusing!

This is a classic test-maker's trap! The value $y = 2$ is the *change* in length (how much it squished). But the question asked for the *new total length* of the spring. If it started at 8 meters and squished by 2 meters, its new length is $8 - 2 = 6$ meters.

Q: Do we just ignore the mass of the spring completely? In real life, springs are heavy!

Yes, in standard JEE mechanics problems, we assume the spring is 'ideal' and massless unless stated otherwise. If the spring had mass, some of the ball's kinetic energy would be used to move the coils of the spring itself, which requires much more complex calculus to solve!

Question

A ball of mass $4	ext{ kg}$, moving with a velocity of $10	ext{ ms}^{-1}$, collides with a spring of length $8	ext{ m}$ and force constant $100	ext{ Nm}^{-1}$. The length of the compressed spring is $x	ext{ m}$. The value of $x$ to the nearest integer, is ............ .

Sigma Percentile · Accepted Answer

### The Setup: A Rolling Ball and a Waiting Spring

Imagine you are standing in a physics lab, watching a heavy ball of mass $m = 4	ext{ kg}$ rolling smoothly across a frictionless floor. It's moving with a brisk velocity of $v = 10	ext{ m/s}$.

Directly in its path is a relaxed spring attached firmly to a wall. The spring has an original, uncompressed length of $L = 8	ext{ m}$ and a stiffness, or force constant, of $k = 100	ext{ N/m}$. The stage is set for a classic collision. What exactly happens when the ball makes contact with the spring?

### The Physics: A Perfect Energy Exchange

As soon as the ball hits the spring, it doesn't just stop instantly. It pushes against the spring, and the spring pushes back. The ball begins to slow down, and the spring begins to compress.

At the exact moment of **maximum compression**, the ball comes to a momentary halt before it gets pushed back. Because we are dealing with an ideal, frictionless system, no energy is lost to heat or sound. The entire **Kinetic Energy** of the moving ball is perfectly converted into the **Elastic Potential Energy** stored within the compressed spring.

We can write this beautiful conservation of mechanical energy mathematically as:

$$\frac{1}{2}mv^2 = \frac{1}{2}ky^2$$

Here, $y$ represents the maximum distance the spring has been compressed from its natural length.

### The Math: Finding the Compression

Now, let's plug our known values into the master equation. We know the mass $m = 4$, the velocity $v = 10$, and the spring constant $k = 100$.

$$\frac{1}{2}(4)(10)^2 = \frac{1}{2}(100)y^2$$

We can immediately cancel out the $\frac{1}{2}$ on both sides to simplify our lives. Squaring the velocity gives us $100$.

$$4 	imes 100 = 100 	imes y^2$$

$$400 = 100y^2$$

Dividing both sides by $100$, we get:

$$y^2 = 4$$

Taking the square root, we find that the maximum compression is:

$$y = 2	ext{ m}$$

### The Final Catch: Reading the Question Carefully

It is incredibly tempting to stop here and declare $2$ as the answer. But wait! This is where many students fall into a classic trap.

The question does **not** ask for the compression of the spring. It specifically asks for the **length of the compressed spring**, denoted as $x$.

To find the final length, we must subtract the compression $y$ from the original natural length $L$:

$$x = L - y$$

$$x = 8 - 2$$

$$x = 6	ext{ m}$$

And there we have it! The final length of the compressed spring is exactly $6	ext{ m}$. Always remember to double-check what the question is actually asking for before you bubble in that final answer!

Enter Numerical Value:

Visualized Solution

$Visualizing the Collision$

$Conservation of Mechanical Energy$

$Substituting the Values$

$Calculating Compression (y)$

$Finding the Final Length (x)$

$What Happens Next?$

The Sigma Insight: Conservation of Mechanical Energy

The Setup

The Physics

The Math

The Final Catch

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