Question

A bullet of mass 10gm moving with a speed of 500 m/s gets embedded in a tree after penetrating
5cm into it. Calculate the average retarding force exerted by the wood on the bullet and the work
done by the wood in bringing the bullet to stop.
(a) 25 N, 12.50 joule
(b) 250 N, 1250 joule
(c) 25 KN, 1.250 joule
(d) 25 KN, 1250 joule

Answer 1

Answer:

25 kN, 1250 joule

Explanation:

It is given that,

Mass of the bullet, m = 10 gm = 0.01 kg

Initial speed of the bullet, u = 500 m/s

Final speed of the bullet, v = 0

Distance covered by the bullet after penetrating, d = 5 cm = 0.05 m

using third equation of motion :

$v^2-u^2=2ad$

$0-(500)^2=2a\times 0.05$

$a=-25\times 10^5\ m/s^2$

Retarding force, F = m a

$F=0.01\ kg\times 25\times 10^5\ m/s^2$

$F=25\times 10^3\ N$

So, F = 25 kN

So, the average retarding force exerted by the wood on the bullet is 25 kN.

Work  done by the wood in bringing the bullet to stop, W = F × d

W = 25 kN × 0.05 m

W = 1250 Joule

Hence, the correct option is (d)