A ball of mass 0.2 kg rests on a vertical post of height 5 m. A bullet of mass 0.01 kg, travelling w

Q: Why do both the ball and the bullet take the same time to hit the ground?

Because gravity acts equally on all objects regardless of their mass or horizontal velocity. Since both start with zero vertical velocity and fall the same vertical distance ($5\text{ m}$), they hit the ground simultaneously.

Q: Why is horizontal momentum conserved during the collision?

The collision occurs over an extremely short time interval. During this split second, the external horizontal forces (like air resistance) are negligible compared to the massive internal impact forces between the bullet and the ball.

Q: Why don't we use conservation of kinetic energy here?

The problem states that the bullet hits the ball and they travel independently, but it doesn't say the collision is perfectly elastic. In fact, a bullet passing through or embedding into a ball usually involves deformation and heat, meaning kinetic energy is lost.

Q: What if the bullet had embedded itself in the ball?

If the bullet embedded itself, it would be a perfectly inelastic collision. They would move together with a common final velocity, and we would use $m_{\text{bullet}} v = (m_{\text{ball}} + m_{\text{bullet}}) v_{\text{final}}$.

Q: Does the height of the post affect the final answer?

Yes! The height determines the time of flight ($t = \sqrt{2h/g}$). A taller post would mean more time in the air, which would change the calculated post-collision velocities for the given ranges, ultimately changing the required initial velocity of the bullet.

Question

A ball of mass $0.2	ext{ kg}$ rests on a vertical post of height $5	ext{ m}$. A bullet of mass $0.01	ext{ kg}$, travelling with a velocity $v	ext{ m/s}$ in a horizontal direction, hits the centre of the ball. After the collision, the ball and bullet travel independently. The ball hits the ground at a distance of $20	ext{ m}$ and the bullet at a distance of $100	ext{ m}$ from the foot of the post. The initial velocity $v$ of the bullet is

Options:

(A) $250	ext{ m/s}$

(B) $250\sqrt{2}	ext{ m/s}$

(C) $400	ext{ m/s}$

(D) $500	ext{ m/s}$

Sigma Percentile · Accepted Answer

### The Setup: A Split-Second Encounter

Imagine a serene scene: a ball resting peacefully atop a $5	ext{ m}$ high vertical post. Suddenly, the tranquility is shattered as a bullet, traveling horizontally at a blazing speed $v$, strikes the ball. In a fraction of a second, the collision is over, and both the ball and the bullet are sent flying through the air as horizontal projectiles.

Our mission is to act as forensic physicists. By examining where the ball and the bullet land, we need to rewind time and determine the initial velocity of the bullet. This problem is a beautiful symphony of two fundamental physics concepts: **Projectile Motion** and the **Conservation of Linear Momentum**.

### The Fall: Gravity is the Great Equalizer

The moment the ball and the bullet leave the post, they become projectiles. Because the bullet struck horizontally, neither object has any initial vertical velocity ($u_y = 0$).

This is where gravity takes over. Regardless of how fast they are moving horizontally, gravity pulls them down at the exact same rate. The time it takes for any horizontal projectile to hit the ground depends only on the height from which it falls. We can find this time of flight using the kinematic equation:

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

Plugging in our values ($h = 5	ext{ m}$ and $g = 10	ext{ m/s}^2$):

$$t = \sqrt{\frac{2 	imes 5}{10}} = 1	ext{ s}$$

Both the ball and the bullet are in the air for exactly **one second**. This is a crucial piece of the puzzle!

### The Flight: Tracing the Horizontal Paths

Now that we know they were flying for $1	ext{ s}$, we can look at their horizontal ranges to figure out how fast they were launched. The horizontal velocity of a projectile remains constant throughout its flight because there are no horizontal forces acting on it (ignoring air resistance).

The relationship is simple: $	ext{Range} = 	ext{Horizontal Velocity} 	imes 	ext{Time}$.

For the ball, it landed $20	ext{ m}$ away:
$$20 = v_{	ext{ball}} 	imes 1 \implies v_{	ext{ball}} = 20	ext{ m/s}$$

For the bullet, it traveled much further, landing $100	ext{ m}$ away:
$$100 = v_{	ext{bullet}} 	imes 1 \implies v_{	ext{bullet}} = 100	ext{ m/s}$$

We have successfully deduced the velocities of both objects immediately *after* the collision!

### The Impact: Conservation of Momentum

Now, we rewind to the split second of the impact. The collision between the bullet and the ball involves massive internal forces, but the external horizontal forces are practically zero. Therefore, the total horizontal momentum of the system is conserved.

The momentum before the collision must equal the momentum after the collision:
$$p_{	ext{initial}} = p_{	ext{final}}$$

Before the impact, only the bullet is moving. After the impact, both are moving. Let's set up the equation using their masses ($m_{	ext{bullet}} = 0.01	ext{ kg}$, $m_{	ext{ball}} = 0.2	ext{ kg}$):

$$m_{	ext{bullet}} v = m_{	ext{ball}} v_{	ext{ball}} + m_{	ext{bullet}} v_{	ext{bullet}}$$

### The Final Calculation

Let's substitute the values we've found into our momentum equation:

$$0.01 	imes v = (0.2 	imes 20) + (0.01 	imes 100)$$

The math simplifies beautifully:

$$0.01 v = 4 + 1$$
$$0.01 v = 5$$

To find the initial velocity $v$, we simply divide by $0.01$:

$$v = \frac{5}{0.01} = 500	ext{ m/s}$$

The bullet was traveling at a staggering **$500	ext{ m/s}$** before it struck the ball. By blending the kinematics of falling bodies with the dynamics of collisions, we've solved the mystery!

Select Answer:

Visualized Solution

Visualizing the Setup

Time of Flight Formula

Substituting Values for Time

Calculating Time of Flight

Horizontal Range Formula

Setting up Ball's Range

Calculating Ball's Velocity

Setting up Bullet's Range

Calculating Bullet's Velocity

Conservation of Momentum

Setting up Momentum Equation

Executing the Multiplication

Total Final Momentum

Final Answer

The Sigma Insight: Conservation of Linear Momentum

The Setup

The Fall

The Flight

The Impact

The Final Calculation

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