Answer:
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
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
