A bullet of 10 g, moving with velocity v, collides head-on with the stationary bob of a pendulum and

Q: Why do we use $\sqrt{5gl}$ instead of $\sqrt{2gl}$ for the bob's velocity?

$\sqrt{2gl}$ is only enough for the bob to reach the horizontal position (90 degrees). To complete a full vertical circle, the string must not go slack at the highest point, which requires a minimum velocity of $\sqrt{5gl}$ at the bottom.

Q: Why is momentum conserved but not kinetic energy?

The collision between the bullet and the bob is inelastic. Energy is lost to heat and deformation. However, because there are no external horizontal forces acting on the system during the brief impact, horizontal momentum is strictly conserved.

Q: Why did we take the recoil velocity as negative?

Velocity is a vector quantity, meaning direction matters. If we define the bullet's initial forward direction as positive, then bouncing backwards (recoiling) must be represented with a negative sign.

Q: Does the tension in the string affect the conservation of momentum?

No. The tension acts in the vertical direction, while the collision and the resulting momentum exchange happen entirely in the horizontal direction. Vertical forces do not affect horizontal momentum.

Q: What would happen if the bullet got embedded in the bob?

If the bullet got embedded, it would be a perfectly inelastic collision. Both the bullet and the bob would move together with a common final velocity $v'$. The momentum equation would change to $mv = (m + M)v'$.

Question

A bullet of $10	ext{ g}$, moving with velocity $v$, collides head-on with the stationary bob of a pendulum and recoils with velocity $100	ext{ m/s}$. The length of the pendulum is $0.5	ext{ m}$ and mass of the bob is $1	ext{ kg}$. The minimum value of $v$ .......... $	ext{m/s}$, so that the pendulum describes a circle. (Assume, the string to be inextensible and $g = 10	ext{ m/s}^2$)

Sigma Percentile · Accepted Answer

### The Setup: A Bullet Meets a Bob

Imagine you are standing in a physics lab, observing a classic ballistic pendulum experiment. But there is a twist!

Instead of the bullet getting embedded in the wooden bob, it strikes the bob and violently recoils backwards.

We are given a bullet of mass $m = 10 	ext{ g}$ (which is $0.01 	ext{ kg}$) moving with an unknown initial velocity $v$. It hits a stationary pendulum bob of mass $M = 1 	ext{ kg}$ suspended by a string of length $l = 0.5 	ext{ m}$.

After the collision, the bullet bounces back with a recoil velocity of $v_b = 100 	ext{ m/s}$. Our mission is to find the minimum initial velocity $v$ required so that the bob completes a full vertical circle.

### The Vertical Circle Constraint

Before we analyze the collision, we must understand what the bob needs to achieve. The problem states that the pendulum must describe a full circle.

For a mass attached to a string to just complete a vertical circle, it must have enough kinetic energy at the bottom to reach the top without the string going slack.

The critical condition for this is that the velocity at the lowest point must be at least:

$$v' = \sqrt{5gl}$$

Let's calculate this required velocity for our bob. We know the acceleration due to gravity $g = 10 	ext{ m/s}^2$ and the length of the string $l = 0.5 	ext{ m}$.

$$v' = \sqrt{5 	imes 10 	imes 0.5}$$

$$v' = \sqrt{25} = 5 	ext{ m/s}$$

So, the bob must shoot forward with a velocity of exactly $5 	ext{ m/s}$ immediately after the impact.

### Conservation of Momentum: The Collision

Now, let's rewind to the exact moment of the collision. The impact happens in a fraction of a second. During this infinitesimal time, the horizontal forces (like air resistance) are negligible.

Because there is no net external horizontal force, the **Law of Conservation of Linear Momentum** holds perfectly true.

The total momentum before the collision must equal the total momentum after the collision:

$$p_{	ext{initial}} = p_{	ext{final}}$$

Initially, only the bullet is moving. The bob is at rest. So, the initial momentum is simply $m 	imes v$.

After the collision, both objects are moving. But we must be extremely careful with our sign convention! Velocity is a vector. Since the bullet recoils backwards, its final velocity must be negative.

$$m v + M(0) = m(-v_b) + M v'$$

### The Final Calculation

Let's substitute all our known values into the momentum equation. Remember to use standard SI units (kilograms and meters per second).

$$0.01 	imes v = 0.01 	imes (-100) + 1 	imes 5$$

Now, we perform the arithmetic. Multiplying $0.01$ by $-100$ gives us $-1$.

$$0.01 v = -1 + 5$$

$$0.01 v = 4$$

To isolate $v$, we divide $4$ by $0.01$:

$$v = \frac{4}{0.01} = 400 	ext{ m/s}$$

And there we have it! The bullet must be fired with a minimum initial velocity of $400 	ext{ m/s}$ to ensure the pendulum bob swings all the way around the circle.

Physics is beautiful when you break it down step by step!

Enter Numerical Value:

Visualized Solution

$Initial Setup$

$Condition for Vertical Circle$

$Calculating v^{'}$

$Conservation of Momentum$

$Substituting Values$

$Simplifying the Equation$

$Final Calculation$

$The Way Forward$

The Sigma Insight: Vertical Circular Motion

The Setup

The Vertical Circle Constraint

Conservation of Momentum

The Final Calculation

Similar Questions

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Enter Numerical Value:

Visualized Solution

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

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Condition for Vertical Circle

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

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

.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; } Simplifying the Equation

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

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