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
Verified Correct Answer: E.
Physical Principles & Numerical Solution
F = ma
= m\(\frac{(v^2 - u^2)}{2s}\)
F = 0.05\(\frac{(0 - (500)^2)}{2 \times 0.25}\)
F = 25 000N
Source: Myschool Past Questions Archive (Record #28565)