Question

A bullet fired into a wall looses half of its velocity after penetrating 3 cm. Further it can penetrate a maximum of
A. 3 cm B. 2 cm C. 1 cm D. 0.5 cm

Answer 1

Let the initial velocity of the bullet be v and final velocity of the object should be v/2.

Now, the bullet is able to penetrate 3cm into the wall. » s = 3cm.

From Third Kinematic Equation,

$\sf \: {v}^{2} - {u}^{2} = 2as \\ \\ \longrightarrow \sf \: \big( \dfrac{v}{2} \big) {}^{2} = {v}^{2} + 2a(3) \\ \\ \longrightarrow \sf \: - \dfrac{3 {v}^{2} }{4} = 6a\\ \\ \longrightarrow \sf \: - \dfrac{ {3v}^{2} }{24 } = a$

If the bullet was to travel 'x' distance more,

$\longrightarrow \sf \: 0 {}^{2} - \dfrac{ {v}^{2} }{4} = 2 \times - \dfrac{3 {v}^{2} }{24} x \\ \\ \longrightarrow \sf - \dfrac{v {}^{2} }{4} = - \dfrac{v {}^{2} }{4} x \\ \\ \longrightarrow \boxed{ \boxed{\sf \: x = 1cm}}$

Option (c) is correct.

Answer 2

let the friction force acting on the bullet is F.

Loss of kinetic energy = Work done by friction

$\frac{1}{2} mv {}^{2} - ( \frac{1}{2} m( \frac{v}{2} ) {}^{2} ) = fx$

$\frac{3}{8} mv {}^{2} = f \times 3 \times 10 {}^{ - 2}$

$f = \frac{( \frac{1}{2}mv {}^{2})}{ (\frac{4}{3} \times 3 \times 10 {}^{ - 2} )} = \frac{ (\frac{1}{2} mv {}^{2}) }{4 \times 10 {}^{ - 2} }$

if total Kinetic energy is lost;

$\dfrac{1}{2}$ mv²=fx$\tiny{2}$

$( \frac{1}{2} mv {}^{2} ) = \frac{ (\frac{1}{2}mv {}^{2} ) }{4 \times 10 {}^{ - 2} } x {}^{2}$

$x \tiny{2}$= $4 \times 10 {}^{ - 2} = 4cm$

Hence,

bullet penetrates further =

x$\tiny{2}$−x=4−3=1cm