Bullet Question

Question

A bullet is shot out of a gun with a velocity of 33.0 m/s {right} and maintains that velocity as a frictionless projectile until it hits a large target. It enters a barn-size structure which is made up of a large layer of cardboard followed by a large layer of rubber. The bullet slows down with an acceleration of 10.0 m/s2 {left} as it passes through the cardboard layer in exactly 1.00 s. Once it enters the rubber layer, it comes to rest after 14.0 m. INCLUDE diagram and…

i) calculate how deep (long) the cardboard layer was

ii) calculate how fast the bullet was moving when it entered the rubber

iii) calculate the acceleration while in the rubber

iv) calculate how much time the bullet was moving in the rubber

Answer 1

Let

v₀ = 33 m/s be the velocity before entering the cardboard,

a₁ = -10 m/s² be the acceleration inside the cardboard (negative because it is deceleration),

v₁ be the velocity as it exits the cardboard and enters the rubber,

s₁ be the distance travelled inside the cardboard,

t₁ = 1 s be the time spent in the cardboard,

a₂ be the acceleration inside the rubber,

v₂ = 0 be the final velocity inside the rubber,

s₂ = 14 m be the distance travelled inside the rubber,

t₂ be the time spent in the rubber

The above quantities are related by

1) v₁ = v₀ + a₁t₁

2) s₁ = v₀t₁ + a₁t₁²/2

3) v₂ = v₁ + a₂t₂

4) s₂ = v₁t₂ + a₂t₂²/2

Plugging in actual numbers

1) v₁ = 33 - 10×1 = 23 m/s

2) s₁ = 33×1 - 10×1²/2 = 28 m

3) 0 = 23 + a₂t₂

a₂ = -23/t₂

4) 14 = 23t₂ + a₂t₂²/2 = 23t₂ - 23t₂/2 = 23t₂/2

t₂ = 28/23 s

i) The depth of the cardboard is s₁ = 28 m

ii) The bullet entered the rubber with speed v₁ = 23 m/s

iii) Acceleration in the rubber is a₂ = -23/t₂ = -23/(28/23) = -23²/28 = -18.9 m/s²

iv) In the rubber the bullet spent t₂ = 28/23 = 1.2 s