A large block of wood of mass M = 5.99 kg is hanging from two long massless cords. A bullet of mass

Q: Why can't we just use conservation of mechanical energy for the entire process from the start?

Because the collision is perfectly inelastic. When the bullet embeds itself into the wood, a massive amount of kinetic energy is lost to heat, sound, and deformation of the wood. Energy is only conserved *after* the collision during the smooth swing.

Q: Why is momentum conserved during the collision if gravity is acting downwards?

The collision happens almost instantaneously. The impulsive forces between the bullet and the block are so massive that the effect of external forces like gravity during that tiny time interval is negligible.

Q: Does the length of the hanging cords matter in this problem?

No, it doesn't. The problem directly gives you the vertical height $h$ the center of mass rises. If the angle of the swing was given instead, then you would need the length of the cords to calculate $h$ using $h = L(1 - \cos\theta)$.

Q: Why do we add the masses $(M+m)$ after the collision?

Because the bullet gets embedded inside the block. They stick together and move as a single combined object with a shared velocity.

Q: What if the bullet passed completely through the block instead of getting embedded?

If the bullet passed through, it would emerge with some final velocity $v_f$. The momentum equation would change to $mv = Mv_{\text{block}} + mv_f$. You would need to know the bullet's exit velocity to solve for the block's speed.

Question

A large block of wood of mass $M = 5.99	ext{ kg}$ is hanging from two long massless cords. A bullet of mass $m = 10	ext{ g}$ is fired into the block and gets embedded in it. The system (block + bullet) then swing upwards, their centre of mass rising a vertical distance $h = 9.8	ext{ cm}$ before the (block + bullet) pendulum comes momentarily to rest at the end of its arc. The speed of the bullet just before collision is
(Take $g = 9.8	ext{ ms}^{-2}$)

Options:

(A) $841.4	ext{ m/s}$

(B) $811.4	ext{ m/s}$

(C) $831.4	ext{ m/s}$

(D) $821.4	ext{ m/s}$

Sigma Percentile · Accepted Answer

### The Ballistic Pendulum Setup

Imagine a heavy wooden block hanging peacefully from two long cords. Suddenly, a high-speed bullet strikes it, embedding itself deep inside the wood. The sheer impact forces the entire system—the block and the bullet together—to swing upward against gravity. This classic physics setup is known as a **ballistic pendulum**, and it was historically used to measure the speed of projectiles before modern electronics existed.

To solve this, we must break the event into two distinct, non-overlapping phases: the instantaneous collision, and the subsequent graceful swing.

### Phase 1: The Collision (Momentum Conservation)

The collision happens in a fraction of a millisecond. During this tiny time window, external forces like gravity and tension haven't had enough time to significantly alter the system's momentum. Therefore, we can safely apply the **Conservation of Linear Momentum**.

Since the bullet gets embedded in the block, this is a perfectly inelastic collision. The initial momentum of the bullet equals the final momentum of the combined mass:

$$mv = (M + m)v_1$$

Here, $v_1$ is the velocity of the combined mass just after the collision.

### Phase 2: The Swing (Energy Conservation)

Immediately after the collision, the combined mass $(M+m)$ starts moving with velocity $v_1$. As it swings upward, it slows down, converting its kinetic energy entirely into gravitational potential energy until it momentarily stops at a height $h$.

Because the tension in the cords does no work (it's always perpendicular to the motion), mechanical energy is conserved during the swing:

$$K_{	ext{initial}} = U_{	ext{final}}$$

$$\frac{1}{2}(M + m)v_1^2 = (M + m)gh$$

Notice how beautifully the combined mass $(M+m)$ cancels out from both sides! This leaves us with a simple expression for the velocity just after the collision:

$$v_1 = \sqrt{2gh}$$

### The Master Equation and Calculation

Now, we bring this result back to our momentum equation to find the bullet's initial speed $v$:

$$mv = (M + m)\sqrt{2gh}$$

$$v = \left(\frac{M + m}{m}ight)\sqrt{2gh}$$

This is the master equation for any standard ballistic pendulum problem. Before we plug in the numbers, we must ensure all units are in the standard SI system. The bullet's mass must be in kilograms ($10	ext{ g} = 0.01	ext{ kg}$), and the height must be in meters ($9.8	ext{ cm} = 0.098	ext{ m}$).

$$v = \left(\frac{5.99 + 0.01}{0.01}ight)\sqrt{2 	imes 9.8 	imes 0.098}$$

$$v = \left(\frac{6.00}{0.01}ight)\sqrt{2 	imes 9.8 	imes \frac{9.8}{100}}$$

Let's simplify the square root smartly to avoid messy arithmetic:

$$v = 600 	imes \frac{9.8}{10} 	imes \sqrt{2}$$

$$v = 60 	imes 9.8 	imes 1.414$$

$$v = 588 	imes 1.414 \approx 831.4 	ext{ m/s}$$

The bullet was traveling at a staggering **$831.4	ext{ m/s}$** just before impact!

Select Answer:

Visualized Solution

$The Ballistic Pendulum$

$Conservation of Linear Momentum$

$Conservation of Mechanical Energy$

$Velocity Just After Collision$

$The Master Equation for Bullet’s Speed$

$Substituting the Values$

$Calculating the Final Speed$

$Final Answer$

$Food for Thought$

The Sigma Insight: Conservation of Mechanical Energy

The Ballistic Pendulum Setup

Phase 1

Phase 2

The Master Equation and Calculation

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Select Answer:

Visualized Solution

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

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

.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; } Velocity Just After Collision

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The Master Equation for Bullet’s Speed

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

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

.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; } Food for Thought