Question

A bullet when fired at a fixed target has its speed decreased to 50% after penetrating 15 cm into it. What additional thickness it will penetrate before coming to rest?
1. 15 cm
2. 10 cm
3. 5 cm
4. 7.5 cm

Answer 1

Correct Answer - Option 3 : 5 cm

Ans: Option 3)

CONCEPT:

• Work-Energy theorem for a Variable force: The work done by the resultant force acting on a particle is equal to the change in kinetic energy of the particle.

\(W= \frac{1}{2}m×(V^{2}_{2}-V_{1}^{2})\)

Where m = mass and V₁ , V₂  are the initial and final velocities respectively.

CALCULATION:

Given that the velocity decreases to 50% after penetrating 15 cm, i.e.  \(V_{2} \) = \(\frac{V_{1}}{2}\) 

Let x be the distance travelled by the bullet through 15 cm.

and F be the retarding force

Using work-energy theorem 

\(-Fx= \frac{1}{2}m×(V_{2}^{2}-V^{2}_{1})\)

\(-Fx= \frac{1}{2}m×(\frac{V^{2}_{1}}{4}-V^{2}_{1})\)

by solving we get,

 \(\frac{1}{2}mV_1^{2}= \frac{3}{4}Fs\) ........... (A)

Let \(x'\) be the distance travelled by the bullet inside the target when the velocity becomes zero.

i.e for V₂ = 0 

Using work energy theorem we get 

\(-Fx'=\frac{1}{2}m(0^{2}-V_{1}^{2})\)

\(\frac{1}{2}mV_{1}^{2}= Fx'\) ...........(B)

From equation (A) and (B)

\(Fx'=\frac{4}{3}Fx\)

Given x = 15 cm

x^' = \(\frac 43\)× 15

we get x^'= 20 cm

Therefore the additional thickness bullet will penetrate before coming to rest = 20 - 15 = 5 cm 

The correct answer is option 3)