A bullet of mass 15 g has a speed of 400m/s. If the bullet strikes at target brought to rest in 2 cm. Calculate the average net force acting on the bullet.

Question

I did it this way, is it correct??

1) m = 15g,

so m in KG = 15/1000 g

2) u = 400m//s

3) v = 0m/s

4) s = 2cm

so, s in m = 2/100 m

Acc. to Newton's 3rd equation,

v^2 = u^2 + 2as,

0 = 160000 + 2 x a x 2/100

a = -4000000 m/s^2

and F = ma

so F = 15/1000 x -4000000

= - 60000 N or -6 x 10^4 N

Please do answer...

Answer 1

You answered correctly. You performed your analysis using the kinematic equations of motion.

A different way to do this problem is to use energy. The bullet has kinetic energy (E_K) calculated from 1/2mv² and all of that energy is lost in work done on the bullet as it strikes the target (work done by the target on the bullet). That work is F•d.

Since the final bullet has no energy, E_K - W = 0 or E_K = W, meaning 1/2mv² = Fd

dividing both sides by "d", we get:

F = (1/2mv²)/d

F = 1/2(0.015)(400²)/0.02 = 60,000 N and since you define the the velocity of the bullet as positive, then the force would be in the opposite direction so -60,000 N