Answer:
a. The final velocity of the block of mass 2 kg is 3 m/s or 3.86 m/s. The final velocity of the block of mass 1.5 kg is 4 m/s or 2.86 m/s b. The kinetic energy change is 0 J or -12.235 J. Since the collision is elastic, we choose ΔK = 0
Explanation:
From principle of conservation of momentum,
momentum before impact = momentum after impact
Let m₁ = 2 kg, m₂ = 1.5 kg and v₁ = 3 m/s, v₂ = 4 m/s represent the masses and initial velocities of the first and second blocks of mass respectively. Let v₃ and v₄ be the final velocities of the blocks. So,
m₁v₁ + m₂v₂ = m₁v₃ + m₂v₄
(2 × 3 + 1.5 × 4) = 2v₃ + 1.5v₄
6 + 6 = 2v₃ + 1.5v₄
12 = 2v₃ + 1.5v₄
2v₃ + 1.5v₄ = 12 (1)
Since the collision is elastic, kinetic energy is conserved. So
1/2m₁v₁² + 1/2m₂v₂² = 1/2m₁v₃² + 1/2m₂v₄²
1/2 × 2 × 3² + 1/2 × 1.5 × 4² = 1/2 ×2v₃² + 1/2 × 1.5v₄²
9 + 12 = v₃² + 0.75v₄²
21 = v₃² + 0.75v₄²
v₃² + 0.75v₄² = 21 (2)
From (1) v₃ = 6 - 0.75v₄ (3) . Substituting v₃ into (2)
(6 - 0.75v₄)² + 0.75v₄² = 21
36 - 9v₄ + 0.5625v₄² + 0.75v₄² = 21
36 - 9v₄ + 1.3125v₄² - 21 = 0
1.3125v₄² - 9v₄ + 15 = 0
Using the quadratic formula,
v₄ = [-(-9) ± √[(-9)² - 4 × 1.3125 × 15]]/(2 × 1.3125)
= [9 ± √[81 - 78.75]]/2.625
= [9 ± √2.25]/2.625
= [9 ± 1.5]/2.625
= [9 + 1.5]/2.625 or [9 - 1.5]/2.625
= 10.5/2.625 or 7.5/2.625
= 4 m/s or 2.86 m/s
Substititing v₄ into (3)
v₃ = 6 - 0.75v₄ = 6 - 0.75 × 4 = 6 - 3 = 3 m/s
or
v₃ = 6 - 0.75v₄ = 6 - 0.75 × 2.86 = 6 - 2.145 = 3.855 m/s ≅ 3.86 m/s
b. The kinetic energy change ΔK = K₂ - K₁
K₁ = initial kinetic energy of the two blocks = 1/2m₁v₁² + 1/2m₂v₂²
= 1/2 × 2 × 3² + 1/2 × 1.5 × 4² = 9 + 12 = 21 J
K₂ = final kinetic energy of the two blocks = 1/2m₁v₃² + 1/2m₂v₄². Using v = 3 m/s and v = 4 m/s
= 1/2 × 2 × 3² + 1/2 × 1.5 × 4² = 9 + 12 = 21 J.
ΔK = K₂ - K₁ = 21 - 21 = 0
Using v = 3.86 m/s and v = 2.86 m/s
K₂ = 1/2 × 2 × 3.86² + 1/2 × 1.5 × 2.86² = 14.8996 - 6.1347 = 8.7649 J ≅ 8.765 J
ΔK = K₂ - K₁ = 8.765 - 21 = -12.235 J
Since the collision is elastic, we choose ΔK = 0
