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?

25N

50N

250N

500N

25 000N

Correct answer is E

F = ma

= m\(\frac{(v^2 - u^2)}{2s}\)

F = 0.05\(\frac{(0 - (500)^2)}{2 \times 0.25}\)

F = 25 000N

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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?