Question

A 6.5×10−2 kg arrow hits the target at 25 m/s and penetrates 3.8 cm before stopping.
A) What average force did the target exert on the arrow? B)What average force did the arrow exert on the target? C)An identical arrow strikes the target at 65m/s . If the target exerts the same average force as before, what's the penetration depth?

Answer 1

1) We use:
2as = v² - u², with v = 0,
to find the acceleration of the arrow.
2 x 0.038 x a = -(25)²
a = -8.22 x 10³ m/s²
F = ma
F = 6.5 x 10⁻² x -8.22 x 10³ 
F = -534.2 N; the negative direction indicates that the force is in the opposite direction of the motion.

B) The arrows force is the same but in the opposite directioin.
534.2 N

C) a = -8.22 x 10³ m/s²
s = -(65)²/(2 x -8.22 x 10³)
s = 25.7 cm