Two bullets P and Q, masses 10 and 20 g, are moving in the same direction towards a target with velocities of 20 and 10 m/s respectively. Which one of the bullets will pierce a greater distance through the target?

Options

A. p
B. Q
C. Both will cover the same distance
D. Nothing can be decided

Answer

To determine which bullet, P or Q, will pierce a greater distance through the target, we can analyze the problem step by step.

### Step 1: Identify the given data
- Mass of bullet P (mP) = 10 g = 0.01 kg (conversion to kg)
- Mass of bullet Q (mQ) = 20 g = 0.02 kg (conversion to kg)
- Velocity of bullet P (vP) = 20 m/s
- Velocity of bullet Q (vQ) = 10 m/s

### Step 2: Understand the scenario
Both bullets are moving towards the same target. Bullet P is moving faster than bullet Q. When they hit the target, they will experience a retarding force that will decelerate them until they come to rest.

### Step 3: Use the equation of motion
We will use the third equation of motion to find the distance each bullet travels into the target before coming to rest:
\[ v^2 = u^2 + 2as \]
Where:
- \( v \) = final velocity (0 m/s, when the bullet comes to rest)
- \( u \) = initial velocity (20 m/s for bullet P and 10 m/s for bullet Q)
- \( a \) = acceleration (negative, since it's a retarding force)
- \( s \) = distance traveled into the target (S1 for bullet P and S2 for bullet Q)

### Step 4: Calculate the distance for bullet P (S1)
For bullet P:
\[ 0 = (20)^2 + 2(-a)S1 \]
\[ 0 = 400 - 2aS1 \]
Rearranging gives:
\[ 2aS1 = 400 \]
\[ S1 = \frac{400}{2a} = \frac{200}{a} \]

### Step 5: Calculate the distance for bullet Q (S2)
For bullet Q:
\[ 0 = (10)^2 + 2(-b)S2 \]
\[ 0 = 100 - 2bS2 \]
Rearranging gives:
\[ 2bS2 = 100 \]
\[ S2 = \frac{100}{2b} = \frac{50}{b} \]

### Step 6: Relate the forces and accelerations
Assuming the target applies the same retarding force \( F \) on both bullets, we can express the accelerations:
- For bullet P: \( a = \frac{F}{mP} = \frac{F}{0.01} \)
- For bullet Q: \( b = \frac{F}{mQ} = \frac{F}{0.02} \)

### Step 7: Substitute the accelerations back into the distance equations
Substituting \( a \) and \( b \) into the distance equations:
\[ S1 = \frac{200}{\frac{F}{0.01}} = \frac{200 \times 0.01}{F} = \frac{2}{F} \]
\[ S2 = \frac{50}{\frac{F}{0.02}} = \frac{50 \times 0.02}{F} = \frac{1}{F} \]

### Step 8: Compare S1 and S2
Now we can compare the distances:
\[ S1 = \frac{2}{F} \]
\[ S2 = \frac{1}{F} \]
Since \( S1 > S2 \), bullet P will pierce a greater distance through the target than bullet Q.

### Conclusion
Therefore, bullet P will pierce a greater distance through the target compared to bullet Q.

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