Question

In a ballistics test, a 25 g bullet traveling horizontally at 1200 m/s goes through a 45 cm thick 350 kg stationary target and emerges with a speed of 1000 m/s. The target is free to slide on a smooth horizontal surface.

(a) How long is the bullet in the target (assume constant acceleration)?
s
What average force does it exert on the target?
N (magnitude only)
(b) What is the target's speed just after the bullet emerges?

Answer 1

Mass of the bullet = m₁ = 25 g = 0.025 kg

Mas of the block = m₂ = 350 kg

Initial speed of the bullet = V₀ = 1200 m/s

Speed of the bullet after coming out of the block = V₁ = 1000 m/s

Distance the bullet travels through the block = Thickness of the block = D = 45 cm = 0.45 m

Acceleration of the bullet = a

V₁² = V₀² + 2aD

(1000)² = (1200)² + 2a(0.45)

a = -488888.89 m/s²

Time period the bullet is in the target = T

V₁ = V₀ + aT

1000 = 1200 + (-488888.89)T

T = 4.09 x 10^-4 sec

Magnitude of the average force the bullet exerts on the target = F

F = |m₁a|

F = |(0.025)(-488888.89)|

F = 1.222 x 10⁴ N

Speed of the block after the bullet emerges = V₂

By conservation of linear momentum,

m₁V₀ = m₁V₁ + m₂V₂

(0.025)(1200) = (0.025)(1000) + (350)V₂

V₂ = 1.43 x 10^-2 m/s

a) Time period the bullet is in the target = 4.09 x 10^-4 sec

Magnitude of the average force the bullet exerts on the target = 1.222 x 10⁴ N

b) Target's speed just after the bullet emerges = 1.43 x 10^-2 m/s