A block of mass 2 kg is traveling in the positive direction at 3 m/s. Another block of mass 1.5 kg, traveling in the same direction at 4 m/s, collides elastically with the first block. Find the final velocities of the blocks. How much kinetic energy did the system lose

Question

Physics

Ean

17 October, 13:23

A block of mass 2 kg is traveling in the positive direction at 3 m/s. Another block of mass 1.5 kg, traveling in the same direction at 4 m/s, collides elastically with the first block. Find the final velocities of the blocks. How much kinetic energy did the system lose

+3

Answers (1)

Know the Answer?

Mohammed Vance · Answer 1

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