JAMB - Physics (1985 - No. 10)
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
Penjelasan
F = ma
= m\(\frac{(v^2 - u^2)}{2s}\)
F = 0.05\(\frac{(0 - (500)^2)}{2 \times 0.25}\)
F = 25 000N
= m\(\frac{(v^2 - u^2)}{2s}\)
F = 0.05\(\frac{(0 - (500)^2)}{2 \times 0.25}\)
F = 25 000N