Problem 22 - Master Resource Book in JEE Main Physics

Question

A bullet when fired at a target with velocity of \(100 \mathrm{~ms}^{-1}\) penetrates \(1 \mathrm{~m}\) into it. If the bullet is fired at a similar target with a thickness \(0.5 \mathrm{~m}\), then it will emerge from it with a velocity of (a) \(50 \sqrt{2} \mathrm{~m} / \mathrm{s}\) (b) \(\frac{50}{\sqrt{2}} \mathrm{~m} / \mathrm{s}\) (c) \(50 \mathrm{~m} / \mathrm{s}\) (d) \(10 \mathrm{~m} / \mathrm{s}\)

Step-by-Step Solution

Answer

The bullet emerges with a velocity of \( 50 \sqrt{2} \text{ ms}^{-1} \).

Step 1: Identify the initial conditions

We have a bullet with an initial velocity of \( v_i = 100 \text{ ms}^{-1} \) which penetrates a \( 1 \text{ m} \) target and comes to a stop. This setup will help us to determine the deceleration (braking force) due to the target resistance.

Step 2: Calculate deceleration in the first scenario

Using the formula \[v_f^2 = v_i^2 + 2a s\]where \( v_f = 0 \text{ ms}^{-1} \) (final velocity), \( v_i = 100 \text{ ms}^{-1} \) is the initial velocity, \( s = 1 \text{ m} \) is the distance, and \( a \) is the acceleration (or deceleration).Rearranging for deceleration:\[a = \frac{-v_i^2}{2s} = \frac{-(100)^2}{2 \times 1} = -5000 \text{ ms}^{-2}\]

Step 3: Apply deceleration to the new scenario

Now, for the second scenario where the thickness \( s = 0.5 \text{ m} \). Let the final velocity beyond this target be \( v_f \). Using the same deceleration:\[v_f^2 = v_i^2 + 2a s\]Substitute the known values:\[v_f^2 = (100)^2 + 2(-5000)(0.5)\]\[v_f^2 = 10000 - 5000\]\[v_f^2 = 5000\]Solving for \( v_f \):\[v_f = \sqrt{5000} = 50\sqrt{2} \text{ ms}^{-1}\]

Step 4: Conclude the final velocity

The bullet emerges with a velocity of \( 50 \sqrt{2} \text{ ms}^{-1} \). This matches the option (a) given in the question.

Key Concepts

Projectile Motion

Projectile motion refers to the motion of an object thrown or projected into the air, subject to only the acceleration of gravity. An important aspect of projectile motion is that it follows a curved path due to the two-dimensional movement. For Firing bullets like the one in our exercise, the trajectory might be close to straight because of its initial high velocity and air resistance playing a minor role.

While dealing with bullets, the vertical motion due to gravity is often negligible over short distances. The primary factors we consider are initial velocity, the path's length, and the forces that come into play when the bullet penetrates a target material, like deceleration. Hence, this becomes focused on understanding force and material resistance rather than elevation or arc trajectories, more common with larger, slower projectiles.

Deceleration

Deceleration is the process of an object slowing down. In physics, deceleration is typically treated as negative acceleration. When a bullet penetrates a target, deceleration results from the resistance encountered. At the moment of impact, the bullet experiences a sudden force that reduces its velocity as it travels through the material.

In the exercise, once the initial velocity and the target's distance are known, deceleration can be calculated using the kinematic equations. This is necessary for understanding how thick different targets will influence the endpoint speed of the bullet. The formula used for deceleration, derived from the exercise, is \[ a = \frac{-v_i^2}{2s} \]and it allows us to quantify the braking force the material exerts on the bullet as it penetrates through it.

Final Velocity Calculation

Final velocity is an important concept in projectile and penetration analysis. It refers to the speed of a projectile once it has traveled through or emerged from an opposing force like a target. Using the initial speed, deceleration, and distance, we can determine this value mathematically.

In the bullet exercise, we determine the final velocity by rearranging the kinematic equation for motion, \[v_f^2 = v_i^2 + 2a s\].This equation accounts for the change in velocity caused by the resistance encountered in the target. With known deceleration and initial speed values, we find the final velocity of the bullet after passing through the target. Through calculation, we discover the bullet's final velocity emerges at \[v_f = 50\sqrt{2} \text{ ms}^{-1}\].This shows how significantly a thinner target affects the bullet's residual speed.

Kinematic Equations

Kinematic equations are fundamental in understanding the physics of motion. These equations describe relationships between velocity, acceleration, time, and displacement. For scenarios involving constant acceleration, these equations are crucial tools.

The bullet problem utilizes one of the kinematic equations:\[v_f^2 = v_i^2 + 2a s \], which allows us to bridge the initial and final velocities given a known acceleration (or deceleration, in the case of resistance) over a measured distance. Generally, these equations empower us to solve for unknown variables and are widely applicable in various contexts. Whether it's finding how long a projectile is in air or calculating a bullet's exit speed, kinematic equations provide structured methods for equation solving in motion analysis.