A bullet fired at a wooden block of thickness 0.15m manages to penetrate the block. If the mass of the bullet is 0.025kg and the average resisting force of the wood is 7.5 x 103N, calculate the speed of the bullet just before it hits the wooden block.

A.

450ms-1

B.

400ms-1

C.

300ms-1

D.

250ms-1

Correct answer is C

F = ma

Therefore, a = \(\frac{F}{m}\)

a = \(\frac{7.5 \times 10^{3}}{0.025}\)

a = 3 x 105m/s2

using V2 = U2 + 2ax;

V2 = 02 + (2 x 3 x 105 x 0.15)

Therefore V = 300 m/s