Answer:
11,000 kg
(a) 11.2 m/s
(b) 1.6 m/s
Explanation:
Momentum is conserved.
m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂
(2200 kg) (60.0 km/h) + m (0 km/h) = (2200 kg) (10 km/h) + m (10 km/h)
132,000 kg km/h = 22,000 kg km/h + m (10 km/h)
110,000 kg km/h = m (10 km/h)
m = 11,000 kg
Momentum is conserved.
m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂
(m) (-v) + (2m) (5v) = (m) (v₁) + (2m) (v₂)
-mv + 10mv = m v₁ + 2m v₂
9mv = m (v₁ + 2 v₂)
9v = v₁ + 2 v₂
Since the collision is elastic, kinetic energy is also conserved.
½ m₁u₁² + ½ m₂u₂² = ½ m₁v₁² + ½ m₂v₂²
m₁u₁² + m₂u₂² = m₁v₁² + m₂v₂²
(m) (-v)² + (2m) (5v)² = m v₁² + (2m) v₂²
mv² + 50mv² = m v₁² + 2m v₂²
51mv² = m (v₁² + 2 v₂²)
51v² = v₁² + 2 v₂²
We know v = 1.60 m/s. So the two equations are:
14.4 = v₁ + 2 v₂
130.56 = v₁² + 2 v₂²
Solve the system of equations using substitution.
130.56 = (14.4 − 2 v₂)² + 2 v₂²
130.56 = 207.36 − 57.6 v₂ + 4 v₂² + 2 v₂²
0 = 6 v₂² − 57.6 v₂ + 76.8
0 = v₂² − 9.6 v₂ + 12.8
v₂ = [ 9.6 ± √(9.6² − 4(1)(12.8)) ] / 2(1)
v₂ = 1.6 or 8
If v₂ = 1.6 m/s, then v₁ = 14.4 − 2 v₂ = 11.2 m/s.
If v₂ = 8 m/s, then v₁ = 14.4 − 2 v₂ = -1.6 m/s.
We know v₁ can't be -1.6 m/s, since that would mean puck A didn't change speeds after the collision. Therefore, v₁ = 11.2 m/s and v₂ = 1.6 m/s.
