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
The correct answer is: C
Explanation
F = maTherefore, 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