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A 0.05kg bullet travelling at 500ms-1 horizontall...

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?

A.

25N

B.

50N

C.

250N

D.

500N

E.

25 000N

Correct answer is E

F = ma

= m(v2u2)2s

F = 0.05(0(500)2)2×0.25

F = 25 000N