A 20 g bullet pierces through a plate of mass M1 = 1 kg and then comes to rest inside a second plate

Q: Why is momentum conserved when the bullet pierces the plate?

Momentum is conserved because there are no external horizontal forces acting on the system. The force between the bullet and the plate is an internal force, and the table is smooth (frictionless).

Q: Why can't we just use conservation of kinetic energy?

Both collisions are inelastic. When the bullet pierces the first plate and embeds in the second, a significant amount of kinetic energy is lost to heat, sound, and deformation of the materials.

Q: Does the bullet lose mass when piercing the first plate?

In reality, it might, but the problem explicitly instructs us to 'Neglect any loss of material of the plates due to the action of bullet.' We assume the bullet's mass $m$ remains constant.

Q: Why do both plates have the same velocity $v_1$?

This is a specific condition given in the problem statement: 'It is found that the two plates initially at rest, now move with equal velocities.' We use this fact to link the two collision equations.

Q: What does 'percentage loss' mean mathematically?

Percentage loss is calculated as the change in a value divided by its initial value, multiplied by 100. Here, it is $\frac{v - v_2}{v} \times 100\%$, which simplifies to $(1 - \frac{v_2}{v}) \times 100\%$.

Question

A $20	ext{ g}$ bullet pierces through a plate of mass $M_1 = 1	ext{ kg}$ and then comes to rest inside a second plate of mass $M_2 = 2.98	ext{ kg}$ as shown in the figure. It is found that the two plates initially at rest, now move with equal velocities. Find the percentage loss in the initial velocity of the bullet when it is between $M_1$ and $M_2$. Neglect any loss of material of the plates due to the action of bullet. Both plates are lying on smooth table.

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing in a ballistics lab. On a perfectly smooth, frictionless table, two heavy metal plates are resting peacefully.

Plate 1 has a mass of $M_1 = 1	ext{ kg}$, and Plate 2 has a mass of $M_2 = 2.98	ext{ kg}$.

Suddenly, a bullet of mass $m = 20	ext{ g}$ (which is $0.02	ext{ kg}$) is fired horizontally with an initial velocity $v$.

The bullet pierces right through the first plate, emerging with a reduced velocity $v_2$. It then strikes the second plate and gets completely embedded inside it.

The problem gives us a beautiful constraint: after all this chaos, both plates end up moving with the exact same final velocity, $v_1$. Our mission is to find out what percentage of its initial velocity the bullet lost while passing through the first plate.

### The First Collision: Piercing Plate 1

Let's isolate the first event. The bullet hits Plate 1 and passes through it. Because the table is smooth, there are no external horizontal forces acting on the system of (Bullet + Plate 1).

This means we can safely apply the Law of Conservation of Linear Momentum.

Before the collision, only the bullet is moving. The initial momentum is simply $mv$.

After the collision, Plate 1 is moving with velocity $v_1$, and the bullet continues forward with velocity $v_2$. The final momentum of this system is $M_1 v_1 + m v_2$.

Equating the two, we get our first master equation:

$$mv = M_1 v_1 + m v_2$$

### The Second Collision: Embedding in Plate 2

Now, let's look at the second event. The bullet, now traveling at velocity $v_2$, strikes Plate 2 and gets embedded. This is a perfectly inelastic collision.

Again, horizontal momentum is conserved. The initial momentum for this specific event is just the momentum of the bullet, $m v_2$.

After the collision, the bullet and Plate 2 move together as a single combined mass $(M_2 + m)$. The problem states that this combined mass moves with the same velocity $v_1$ as the first plate.

This gives us our second master equation:

$$m v_2 = (M_2 + m) v_1$$

### The Master Equation

We have a system of two equations. Our ultimate goal is to find the percentage loss in the bullet's velocity, which mathematically translates to finding the ratio $\frac{v_2}{v}$.

To do this, we need to eliminate the intermediate variable $v_1$. From our second equation, we can easily isolate $v_1$:

$$v_1 = \frac{m v_2}{M_2 + m}$$

Now, we substitute this expression for $v_1$ back into our first equation. This is where the algebra gets interesting:

$$mv = M_1 \left( \frac{m v_2}{M_2 + m} ight) + m v_2$$

Notice that the term $m v_2$ appears in both parts of the right-hand side. Let's factor it out to clean up the expression:

$$mv = m v_2 \left( \frac{M_1}{M_2 + m} + 1 ight)$$

By taking a common denominator inside the bracket, we can combine the terms:

$$mv = m v_2 \left( \frac{M_1 + M_2 + m}{M_2 + m} ight)$$

### Final Calculation

Look at the elegance of this equation! The mass of the bullet, $m$, appears as a multiplier on both sides. We can cancel it out, leaving us with a direct relationship between the initial and intermediate velocities:

$$\frac{v_2}{v} = \frac{M_2 + m}{M_1 + M_2 + m}$$

Now, it is time to plug in the numerical values given in the problem.

The mass of the bullet is $m = 0.02	ext{ kg}$. 
The mass of Plate 1 is $M_1 = 1	ext{ kg}$. 
The mass of Plate 2 is $M_2 = 2.98	ext{ kg}$.

Let's calculate the numerator and denominator:

Numerator: $M_2 + m = 2.98 + 0.02 = 3.00	ext{ kg}$
Denominator: $M_1 + M_2 + m = 1 + 2.98 + 0.02 = 4.00	ext{ kg}$

Substituting these back into our ratio:

$$\frac{v_2}{v} = \frac{3.00}{4.00} = 0.75$$

This result tells us that the bullet retains $75\%$ of its initial velocity after piercing the first plate.

If it retains $75\%$, then the percentage loss is simply:

$$	ext{Percentage Loss} = 100\% - 75\% = 25\%$$

And there we have it! The bullet lost exactly $25\%$ of its initial velocity.

Enter Numerical Value:

Visualized Solution

Visualizing the Setup

Given Data

First Collision (Bullet & $M_{1}$ )

Second Collision (Bullet & $M_{2})$

Defining the Goal

Isolating $v_{1}$

Substituting $v_{1}$

Factoring out $m v_{2}$

Simplifying the Bracket

Finding the Ratio

Substituting Values

Calculating the Ratio

Final Percentage Loss

The Sigma Insight: Conservation of Linear Momentum

Analyzing the Setup

The First Collision

The Second Collision

The Master Equation

Final Calculation

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