Answer:
003) 2. is the same as when it started from rest
004) 4. tAB / tA'B' = 1
005) vA/vA' = 1/2
Explanation:
I will do 3, 4, and 5 as examples.
003) Impulse = change in momentum
F Δt = m Δv
The same force and same time interval are exerted on the cart, so it's the same impulse. Therefore, the change in momentum is the same as when the cart started from rest.
004) In each case, the block has the same initial height and the same initial vertical velocity, so the time to land is the same.
005) In the inelastic case:
momentum before = momentum after
M₁ v₁ = (M₁ + m₂) vA
vA = M₁ v₁ / (M₁ + m₂)
In the elastic case:
M₁ v₁ = -M₁ v₁' + m₂ vA'
v₁' = (m₂ vA' − M₁ v₁) / M₁
Energy is also conserved in elastic collisions:
½ M₁ v₁² = ½ M₁ v₁'² + ½ m₂ vA'²
M₁ v₁² = M₁ v₁'² + m₂ vA'²
M₁ v₁² = M₁ (m₂ vA' − M₁ v₁)² / M₁² + m₂ vA'²
M₁ v₁² = (m₂ vA' − M₁ v₁)² / M₁ + m₂ vA'²
M₁² v₁² = (m₂ vA' − M₁ v₁)² + M₁ m₂ vA'²
M₁² v₁² = m₂² vA'² − 2 m₂ M₁ vA' v₁ + M₁² v₁² + M₁ m₂ vA'²
0 = m₂² vA'² − 2 m₂ M₁ vA' v₁ + M₁ m₂ vA'²
0 = m₂ vA' − 2 M₁ v₁ + M₁ vA'
2 M₁ v₁ = (m₂ + M₁) vA'
vA' = 2 M₁ v₁ / (m₂ + M₁)
Therefore:
vA/vA' = 1/2
