Question

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In an elastic head-on collision, a 0.60 kg cart moving at 5.0 m/s [W] collides with a 0.80 kg cart moving at 2.0 m/s [E]. The collision is cushioned by a spring (k = 1200 N/m).
a) Determine the velocity of each cart after the collision.
b) Determine the maximum compression of the spring.

Answer 1

a) 0.60 kg cart has final velocity 3.0 m/s [E]

0.80 kg cart has final velocity 4.0 m/s [W]

b) 0.12 m

Step-by-step explanation:

Take east to be positive.

a) Momentum is conserved.

m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂

(0.60) (-5.0) + (0.80) (2.0) = (0.60) v₁ + (0.80) v₂

-1.4 = 0.6 v₁ + 0.8 v₂

Kinetic energy is conserved in elastic collisions.

½ m₁u₁² + ½ m₂u₂² = ½ m₁v₁² + ½ m₂v₂²

m₁u₁² + m₂u₂² = m₁v₁² + m₂v₂²

(0.60) (-5.0)² + (0.80) (2.0)² = (0.60) v₁² + (0.80) v₂²

18.2 = 0.6 v₁² + 0.8 v₂²

Solve the system of equations.

-1.4 = 0.6 v₁ + 0.8 v₂

-1.4 − 0.6 v₁ = 0.8 v₂

-1.75 − 0.75 v₁ = v₂

18.2 = 0.6 v₁² + 0.8 (-1.75 − 0.75 v₁)²

18.2 = 0.6 v₁² + 0.8 (3.0625 +2.625 v₁ + 0.5625 v₁²)

182 = 6 v₁² + 8 (3.0625 + 2.625 v₁ + 0.5625 v₁²)

182 = 6 v₁² + 24.5 + 21 v₁ + 4.5 v₁²

0 = 10.5 v₁² + 21 v₁ − 157.5

0 = v₁² + 2 v₁ − 15

0 = (v₁ − 3) (v₁ + 5)

v₁ = 3 or -5

Since u₁ = -5.0 m/s, v₁ must be 3.0 m/s.

Solving for v₂:

v₂ = -0.75 v₁ − 1.75

v₂ = -4.0 m/s

b) The compression of the spring is a maximum when the carts have the same velocity.

Momentum is conserved.

m₁u₁ + m₂u₂ = (m₁ + m₂) v

(0.60) (-5.0) + (0.80) (2.0) = (0.60 + 0.80) v

-1.4 = 1.4 v

v = -1.0

Energy is conserved.

½ m₁u₁² + ½ m₂u₂² = ½ (m₁ + m₂) v² + ½ kx²

m₁u₁² + m₂u₂² = (m₁ + m₂) v² + kx²

(0.60) (-5.0)² + (0.80) (2.0)² = (0.60 + 0.80) (-1.0)² + (1200) x²

18.2 = 1.4 + 1200 x²

16.8 = 1200 x²

x² = 0.014

x = 0.12