A bullet of mass 20 g has an initial speed of 1 ms^-1, just before it starts penetrating a mud wall

Q: Why is the acceleration negative?

The force exerted by the mud wall is a resistive force, meaning it acts in the opposite direction to the bullet's motion. This causes the bullet to slow down, resulting in a negative acceleration (deceleration).

Q: Why did we convert the mass from grams to kilograms?

In physics, the standard SI unit for mass is the kilogram (kg). Since the force is given in Newtons (which is $\text{kg}\cdot\text{m/s}^2$), using grams would lead to an incorrect calculation of acceleration.

Q: Can I use the Work-Energy Theorem to solve this?

Absolutely! The work done by the resistive force ($-F \cdot s$) is equal to the change in the bullet's kinetic energy ($\frac{1}{2}mv^2 - \frac{1}{2}mu^2$). Both methods are perfectly valid and will give you the same answer.

Q: What would happen if the wall was 50 cm thick instead of 20 cm?

If the wall were thicker, the resistive force would act over a longer distance. If you calculate $v^2$ for $s = 0.5\text{ m}$, you'll find it becomes negative, which means the bullet would come to a complete stop inside the wall before emerging.

Q: Why didn't we use the first equation of motion ($v = u + at$)?

The first equation requires us to know the time ($t$) the bullet spends inside the wall. Since we are given the displacement ($s$) instead of time, the third equation ($v^2 = u^2 + 2as$) is the most direct and efficient choice.

Question

A bullet of mass $20	ext{ g}$ has an initial speed of $1	ext{ ms}^{-1}$, just before it starts penetrating a mud wall of thickness $20	ext{ cm}$. If the wall offers a mean resistance of $2.5 	imes 10^{-2}	ext{ N}$, the speed of the bullet after emerging from the other side of the wall is close to

Options:

(A) $0.3	ext{ ms}^{-1}$

(B) $0.4	ext{ ms}^{-1}$

(C) $0.1	ext{ ms}^{-1}$

(D) $0.7	ext{ ms}^{-1}$

Sigma Percentile · Accepted Answer

### The Physics of a Speeding Bullet: Penetrating the Mud Wall

Imagine a bullet, with a mass of $20	ext{ g}$, flying towards a mud wall at a speed of $1	ext{ m/s}$. The wall has a thickness of $20	ext{ cm}$. As the bullet penetrates the wall, the mud exerts a resistance force on it. This force will slow the bullet down. We need to find the speed of the bullet right as it emerges from the other side.

### The Resistance: Newton's Second Law

First, we need to determine how much deceleration this resistance force produces. To do this, we will use Newton's second law of motion, $F = ma$.

There is a catch here. Do not make a silly mistake. The mass is given in grams, so we must convert it to kilograms. Also, since the force opposes the motion, we must include a minus sign. So acceleration $a$ equals:

$$a = \frac{-2.5 	imes 10^{-2}	ext{ N}}{20 	imes 10^{-3}	ext{ kg}}$$

Let's calculate this. In the denominator, $20 	imes 10^{-3}$ becomes $2 	imes 10^{-2}$. The $10^{-2}$ terms cancel out. This leaves us with an acceleration $a$ of $-1.25	ext{ m/s}^2$.

### Kinematics: The Third Equation of Motion

Now we have the initial velocity, the displacement, and the acceleration. To find the final velocity, we can use the third equation of kinematics:

$$v^2 - u^2 = 2as$$

Let's substitute the values. The initial velocity $u$ is $1	ext{ m/s}$. The displacement $s$ is $20	ext{ cm}$, which we must convert to meters, giving us $0.2	ext{ m}$. The equation becomes:

$$v^2 - (1)^2 = 2(-1.25)(0.2)$$

Two times $0.2$ is $0.4$. And $0.4$ times $-1.25$ gives $-0.5$. So, $v^2$ equals $1 - 0.5$, which is $0.5$.

### The Final Calculation

If $v^2$ is $0.5$, then $v$ is the square root of $0.5$.

$$v = \sqrt{0.5} \approx 0.707	ext{ m/s}$$

This is approximately **$0.7	ext{ m/s}$**. That is our final answer.

### The Alternative Path: Work-Energy Theorem

Did you get the feel of it? We could have also solved this problem using the Work-Energy theorem. The work done by the resistance force equals the change in kinetic energy:

$$W = \Delta K \implies -F \cdot s = \frac{1}{2}mv^2 - \frac{1}{2}mu^2$$

Try it out yourself, you will get the exact same result! Physics is beautiful because multiple paths often lead to the same truth.

Select Answer:

Visualized Solution

$m = 20 g, u = 1 m/s, s = 20 cm$

$F = - 2.5 \times 1 0^{- 2} N$

$F = ma$

$a = \frac{F}{m}$

$a = - 1.25 m/s^{2}$

$v^{2} - u^{2} = 2 a s$

$v^{2} - (1)^{2} = 2 (- 1.25) (0.2)$

$v^{2} = 0.5$

$v \approx 0.707 m/s$

$W = Δ K$

The Sigma Insight: Newton's Laws of Motion

The Physics of a Speeding Bullet

The Resistance

Kinematics

The Final Calculation

The Alternative Path

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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; } m=20 g,u=1 m/s,s=20 cm

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } F=−2.5×10−2 N

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

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

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } a=−1.25 m/s2

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } v2−u2=2as

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } v2−(1)2=2(−1.25)(0.2)

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } v2=0.5

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } v≈0.707 m/s

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