A 0.05kg bullet travelling at 500ms-1 horizontally strikes a thick vertical wall. It stops after penetrating through the wall horizontal distance of 0.25m. What is the magnitude of the average force the wall exerts on the bullet?
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Correct Answer: Option E
Explanation:
F = ma
= m\(\frac{(v^2 - u^2)}{2s}\)
F = 0.05\(\frac{(0 - (500)^2)}{2 \times 0.25}\)
F = 25 000N
F = ma
= m\(\frac{(v^2 - u^2)}{2s}\)
F = 0.05\(\frac{(0 - (500)^2)}{2 \times 0.25}\)
F = 25 000N