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.

450ms^-1

400ms^-1

300ms^-1

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

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