In a spring gun having spring constant 100 N/m a small ball B of mass 100 g is put in its barrel (as

Q: Why don't we consider the mass of the spring?

In standard physics problems, springs are assumed to be "ideal" or massless unless a specific mass is given. This simplifies the energy conservation equation.

Q: Why is the initial vertical velocity zero?

The problem states the ball leaves the gun "horizontally". This means all of its initial kinetic energy is directed along the x-axis, with no vertical component.

Q: Does the mass of the ball affect the time of flight?

No! The time of flight depends only on the vertical height and gravity ($t = \sqrt{\frac{2H}{g}}$). All objects fall at the same rate in a vacuum, regardless of mass.

Q: Why did we use $g=10$ instead of $9.8$?

Always check the problem statement! It explicitly says "(Take, $g=10\text{m/s}^2$)". Using $9.8$ would lead to a slightly different, and in this context, incorrect answer.

Q: What if there was air resistance?

If air resistance were present, it would oppose the motion, slowing down the horizontal velocity and altering the vertical acceleration. The range $d$ would be shorter, and the trajectory wouldn't be a perfect parabola.

Question

In a spring gun having spring constant $100 	ext{ N/m}$ a small ball $B$ of mass $100 	ext{ g}$ is put in its barrel (as shown in figure) by compressing the spring through $0.05 	ext{ m}$. There should be a box placed at a distance $d$ on the ground, so that the ball falls in it. If the ball leaves the gun horizontally at a height of $2 	ext{ m}$ above the ground. The value of $d$ is ……… m.
(Take, $g=10	ext{m/s}^2$)

Sigma Percentile · Accepted Answer

Have you ever played with a spring-loaded toy gun and wondered exactly where the projectile would land? It feels like pure intuition when you're playing, but behind that simple action lies a beautiful symphony of physics. In this problem, we are going to break down that exact scenario. We have a spring gun, a compressed spring, and a ball that is about to take flight.

This problem is a classic because it perfectly marries two fundamental concepts in physics: **Conservation of Mechanical Energy** and **Projectile Motion**. Let's dive in and see how these two ideas work together to give us the exact landing spot of the ball.

### Analyzing the Setup

Imagine the scene. We are standing $2 	ext{ m}$ above the ground. In our hands (or mounted on a stand), we have a spring gun. The spring inside has a stiffness, or spring constant, of $k = 100 	ext{ N/m}$. We take a small ball of mass $m = 100 	ext{ g}$ (which we must immediately convert to $0.1 	ext{ kg}$ to keep our units consistent) and push it into the barrel, compressing the spring by a distance $x = 0.05 	ext{ m}$.

Right now, the ball is at rest. But it has potential. Specifically, it has **elastic potential energy** stored in that compressed spring. The moment we pull the trigger, the spring will snap back to its natural length, transferring all that stored energy into the ball. The ball will then shoot out horizontally and begin its journey through the air. Our goal is to find exactly where to place a box on the ground so the ball lands perfectly inside it.

### The Master Equation: Energy Conservation

The first phase of our problem happens entirely inside the barrel of the gun. We need to find out how fast the ball is moving the instant it leaves the gun. To do this, we use the principle of Conservation of Mechanical Energy.

Assuming no energy is lost to friction inside the barrel, the elastic potential energy of the spring is completely converted into the kinetic energy of the ball.

The elastic potential energy stored in a compressed spring is given by:
$$U_{	ext{spring}} = \frac{1}{2} k x^2$$

The kinetic energy of the ball as it leaves the spring is:
$$K_{	ext{ball}} = \frac{1}{2} m v^2$$

Equating the two, we get our master equation for this phase:
$$\frac{1}{2} k x^2 = \frac{1}{2} m v^2$$

Notice how the $\frac{1}{2}$ on both sides elegantly cancels out, leaving us with:
$$k x^2 = m v^2$$

Now, let's plug in the numbers we know. The spring constant $k$ is $100$, the compression $x$ is $0.05$, and the mass $m$ is $0.1$:
$$100 	imes (0.05)^2 = 0.1 	imes v^2$$

Squaring $0.05$ gives us $0.0025$. Multiplying that by $100$ shifts the decimal point two places to the right:
$$0.25 = 0.1 	imes v^2$$

Dividing both sides by $0.1$, we find the square of the velocity:
$$v^2 = 2.5$$

Taking the square root, we get the launch velocity:
$$v = \sqrt{2.5} = \sqrt{\frac{25}{10}} = \frac{5}{\sqrt{10}} = 0.5\sqrt{10} 	ext{ m/s}$$

So, the ball leaves the gun horizontally at a speed of $0.5\sqrt{10} 	ext{ m/s}$.

### The Projectile Phase: Time of Flight

The moment the ball leaves the barrel, the spring is no longer pushing it. Now, gravity takes over. Because the ball was fired perfectly horizontally, its initial vertical velocity is zero. It acts as a horizontal projectile.

In projectile motion, the horizontal and vertical motions are completely independent. The time the ball spends in the air is dictated entirely by gravity and the height from which it was dropped.

We can use the second equation of motion for the vertical direction:
$$H = u_y t + \frac{1}{2} g t^2$$

Since the initial vertical velocity $u_y$ is $0$, the equation simplifies beautifully to:
$$H = \frac{1}{2} g t^2$$

Rearranging this to solve for the time of flight $t$, we get:
$$t = \sqrt{\frac{2H}{g}}$$

Let's substitute our height $H = 2 	ext{ m}$ and the acceleration due to gravity $g = 10 	ext{ m/s}^2$:
$$t = \sqrt{\frac{2 	imes 2}{10}} = \sqrt{\frac{4}{10}} = \frac{2}{\sqrt{10}} 	ext{ s}$$

This is the exact amount of time the ball will spend in free fall before it hits the ground.

### Final Calculation: Finding the Range

Now we know how fast the ball is moving horizontally ($v = 0.5\sqrt{10} 	ext{ m/s}$) and how long it will be flying ($t = \frac{2}{\sqrt{10}} 	ext{ s}$).

Because there is no air resistance (as is standard in these problems), the horizontal velocity remains perfectly constant throughout the flight. The horizontal distance, or range $d$, is simply the horizontal speed multiplied by the time of flight:
$$d = v 	imes t$$

Let's bring our two results together:
$$d = \left(0.5\sqrt{10}ight) 	imes \left(\frac{2}{\sqrt{10}}ight)$$

Look at how perfectly the math works out! The $\sqrt{10}$ in the numerator and the $\sqrt{10}$ in the denominator cancel each other out completely. We are left with:
$$d = 0.5 	imes 2$$
$$d = 1 	ext{ m}$$

The box must be placed exactly $1 	ext{ m}$ away from the edge of the gun.

It's incredibly satisfying when a problem that starts with decimals and square roots simplifies down to a clean, whole number. This is the beauty of physics—taking a complex, dynamic situation and using fundamental laws to predict the outcome with absolute precision.

Enter Numerical Value:

Visualized Solution

$Visualizing the Setup$

$Conservation of Mechanical Energy$

$Substituting Values$

$Calculating Launch Velocity$

$Time of Flight$

$Calculating Time$

$Horizontal Range$

$Final Answer$

$The Way Forward$

The Sigma Insight: Conservation of Mechanical Energy

Analyzing the Setup

The Master Equation

The Projectile Phase

Final Calculation

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Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Visualizing the Setup

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Conservation of Mechanical Energy

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Substituting Values

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Calculating Launch Velocity

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Time of Flight

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Calculating Time

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Horizontal Range

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Final Answer

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The Way Forward