A bullet of mass 5 g, travelling with a speed of 210 m/s, strikes a fixed wooden target. One-half of

Q: Why did we ignore the 5g mass of the bullet in the final calculation?

Because the mass $m$ appears in both the kinetic energy formula ($\frac{1}{2}mv^2$) and the heat absorption formula ($ms\Delta T$). When you equate them, the mass cancels out. The temperature rise depends only on the velocity and the material's specific heat.

Q: Why did we multiply the specific heat by 4.2 and 1000?

The specific heat was given in calories per gram. To use it with velocity in meters per second (which yields Joules per kilogram), we must convert calories to Joules (multiply by 4.2) and grams to kilograms (multiply by 1000).

Q: What happens to the other half of the kinetic energy?

The problem states that half the energy goes into the bullet. The other half is absorbed by the wooden target, causing a localized temperature rise in the wood around the impact zone.

Q: Why is it important that the wooden target is 'fixed'?

If the target wasn't fixed, the impact would cause the block to move. By conservation of momentum, some of the initial kinetic energy would be transferred into the macroscopic kinetic energy of the block, leaving less energy to be converted into heat.

Q: Could this temperature rise melt the bullet?

It depends on the bullet's material. A rise of $87.5^\circ\text{C}$ added to room temperature (say $25^\circ\text{C}$) gives a final temperature of about $112.5^\circ\text{C}$. Lead melts at $327.5^\circ\text{C}$, so the bullet would get very hot but would not melt.

Question

A bullet of mass $5	ext{ g}$, travelling with a speed of $210	ext{ m/s}$, strikes a fixed wooden target. One-half of its kinetic energy is converted into heat in the bullet while the other half is converted into heat in the wood. The rise of temperature of the bullet, if the specific heat of its material is $0.030	ext{ cal g}^{-1}	ext{ }^\circ	ext{C}^{-1}$, is close to

Options:

(A) $87.5^\circ	ext{C}$

(B) $83.3^\circ	ext{C}$

(C) $119.2^\circ	ext{C}$

(D) $38.4^\circ	ext{C}$

Sigma Percentile · Accepted Answer

### The Setup: A Speeding Bullet

Imagine a bullet of mass $m = 5	ext{ g}$ tearing through the air at a blistering speed of $v = 210	ext{ m/s}$. It is on a collision course with a fixed wooden target. When it strikes, it doesn't just stop; its immense kinetic energy has to go somewhere. According to the law of conservation of energy, this kinetic energy is transformed into heat.

### The Physics of Impact: Where Does the Energy Go?

The initial kinetic energy of the bullet is given by the classic formula:

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

When the bullet embeds itself into the fixed wooden target, all of this kinetic energy is dissipated as heat. However, the problem provides a crucial constraint: exactly one-half of this generated heat is absorbed by the bullet itself, while the other half is absorbed by the wood. Therefore, the heat energy $\Delta Q$ that goes into raising the bullet's temperature is:

$$\Delta Q = \frac{1}{2} K = \frac{1}{4}mv^2$$

### The Calorimetry Connection

Now, how does this absorbed heat translate into a rise in temperature? We turn to the fundamental principle of calorimetry. The heat absorbed by an object is directly proportional to its mass, its specific heat capacity $s$, and the change in temperature $\Delta T$:

$$\Delta Q = ms\Delta T$$

By equating the two expressions for the heat absorbed by the bullet, we get a beautiful mathematical cancellation:

$$ms\Delta T = \frac{1}{4}mv^2$$

Notice how the mass $m$ appears on both sides? It cancels out completely! This reveals a fascinating physical insight: **the temperature rise of the bullet is entirely independent of its mass**. It depends only on its impact velocity and the material's specific heat capacity. Our master equation simplifies to:

$$\Delta T = \frac{v^2}{4s}$$

### The Trap of Units

Before we rush to plug in the numbers, we must navigate a classic physics trap: inconsistent units. The velocity is given in standard SI units ($	ext{m/s}$), but the specific heat is given in CGS-based units: $s = 0.030	ext{ cal g}^{-1}	ext{ }^\circ	ext{C}^{-1}$.

To use our master equation, we must convert $s$ into standard SI units ($	ext{J kg}^{-1}	ext{ }^\circ	ext{C}^{-1}$):
1. Convert calories to Joules by multiplying by $4.2$.
2. Convert per gram to per kilogram by multiplying by $1000$.

$$s = 0.030 	imes 4.2 	imes 1000 = 126	ext{ J kg}^{-1}	ext{ }^\circ	ext{C}^{-1}$$

### The Final Computation

With our units perfectly aligned, we can now substitute the values into our master equation:

$$\Delta T = \frac{(210)^2}{4 	imes 126}$$

$$\Delta T = \frac{44100}{504} = 87.5^\circ	ext{C}$$

The bullet experiences a sudden temperature spike of **$87.5^\circ	ext{C}$** upon impact. This elegant problem demonstrates how macroscopic kinetic energy seamlessly transitions into microscopic thermal energy.

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Visualized Solution

$I ni t ia l S e t u p$

$K in e t i c E n er g y$

$H e a t C o n v er s i o n$

$C a l or im e t r y P r in c i pl e$

$E q u a t in g E n er g i es$

$U ni t C o n v er s i o n$

$S u b s t i t u t i o n$

$F ina l C a l c u l a t i o n$

$T h e W a y F or w a r d$

The Sigma Insight: Calorimetry

The Setup

The Physics of Impact

The Calorimetry Connection

The Trap of Units

The Final Computation

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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; } Kinetic Energy

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

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

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

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

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

.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