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.
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Correct Answer: Option C
Explanation:
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
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