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: 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