A bullet fired into a fixed wooden block loses half of its velocity after penetrating 40 cm. It comes to rest after penetrating a further distance of

Question

A bullet fired into a fixed wooden block loses half of its velocity after penetrating 40 cm. It comes to rest after penetrating a further distance of (A)  (22/3)cm  (B)  (40/3)cm  (C)  (20/3)cm  (D)  (22/5)cm

Tardigrade Admin · Accepted Answer

Let initial velocity of body at point A is v, AB is 40cm.

From

v^{2} = u^{2} - 2 a s

\Rightarrow

(\frac{v}{2})^{2} = v^{2} - 2 a \times 40

Or

a = \frac{3 v ^{2}}{320}

Let on penetrating 40 cm in a wooden block, the body moves x distance from B to C.
So, for B to C

u = \frac{v}{2}, v = 0

s = x, a = \frac{3 v ^{2}}{320}

(deceleration)

∴

(0)^{2} = (\frac{v}{2})^{2} - 2 \times \frac{3 v ^{2}}{320} \times x

Or

x = \frac{40}{3} c m

Q. A bullet fired into a fixed wooden block loses half of its velocity after penetrating $40 c m$ . It comes to rest after penetrating a further distance of

$\frac{22}{3} c m$

$\frac{40}{3} c m$

$\frac{20}{3} c m$

$\frac{22}{5} c m$

$\frac{26}{5} c m$

Q. A bullet fired into a fixed wooden block loses half of its velocity after penetrating 40cm. It comes to rest after penetrating a further distance of

322​cm

340​cm

320​cm

522​cm