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A bullet of mass 50 g moving with a speed of 100 m s⁻¹ enters a heavy stationary wooden block and stops after penetrating a distance of 50 cm. Estimate the stopping force acting on the bullet (assume that the bullet undergoes constant acceleration within the block).
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m = 50 g = 0.05 kg, u = 100 m s −1 , v = 0, s = 50 cm = 0.5 m.
Using v 2 = u 2 + 2as : 0 = (100) 2 + 2a(0.5) , so a = -10000 1 = -10000 m s -2 .
Force F = ma = 0.05 × (-10000) = -500 N . The stopping force is 500 N, opposite to the bullet's motion.
Stopping force on the bullet
Convert the units
m = 50 g =
0.05 kg
· s = 50 cm =
0.5 m
· u = 100 m s⁻¹ · v = 0
↓
Use
v
2
= u
2
+ 2as
0 = 100
2
+ 2a(0.5)
→
a = -10000
m s
-2
↓
Apply F = ma
F = 0.05 × (-10000)
↓
F = −500 N
— a stopping force of
500 N
, against the motion
