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A 2.5 kg block with a speed of 3.9 m/s collides with a 5.0 kg block that has a speed of 2.6 m/s in the same direction. After the collision, the 5.0 kg block is observed to be traveling in the original direction with a speed of 3.3 m/s. (a) What is the velocity of the 2.5 kg block immediately after the collision? (b) By how much does the total kinetic energy of the system of two blocks change because of the collision? (c) Suppose, instead, that the 5.0 kg block ends up with a speed of 5.2 m/s. What then is the change in the total kinetic energy?

Respuesta :

onyebuchinnaji

Answer:

(a) the velocity of the first block block immediately after the collision is 2.5 m/s  the same direction

(b) change in total kinetic energy of the system is - 0.875 J

(c) change in total kinetic energy of the system is 33.8 J

Explanation:

Given;

mass of first block, m₁ = 2.5 kg

initial speed of first block, u₁ = 3.9 m/s

mass of second block, m₂ = 5.0 kg

initial speed of the second block, u₂ = 2.6 m/s

final speed of the second block, v₂ = 3.3 m/s

Part (A) velocity of the first block block immediately after the collision:

Apply the principle of conservation of linear momentum;

m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂

where;

v₁ is the velocity of the first block block immediately after the collision

(2.5 x 3.9) + (5 x 2.6) = 2.5v₁ + (5 x 3.3)

22.75 = 2.5v₁ + 16.5

2.5v₁ = 22.75  -  16.5

2.5v₁ = 6.25

v₁ = 6.25 / 2.5

v₁ = 2.5 m/s  the same direction

Part (B) change in total kinetic energy of the system:

change in kinetic energy = final kinetic energy  - initial kinetic energy

ΔK =  (¹/₂m₁v₁² + ¹/₂m₂v₂²)  -  (¹/₂m₁u₁² + ¹/₂m₂u₂²)

ΔK =  ( ¹/₂ x 2.5 x 2.5² + ¹/₂ x 5 x 3.3²)  -   (¹/₂ x 2.5 x 3.9² + ¹/₂ x 5 x 2.6²)

ΔK = 35.0375 J  - 35.9125 J

ΔK =  - 0.875 J

Part (C) change in total kinetic energy of the system if second block ends up with a speed of 5.2 m/s

Apply the principle of conservation of linear momentum to determine the final speed of the first block;

m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂

(2.5 x 3.9) + (5 x 2.6) = 2.5v₁ + (5 x 5.2)

22.75 = 2.5v₁ + 26

2.5v₁ = 22.75 - 26

2.5v₁  = - 3.25

v₁ = - 3.25 / 2.5

v₁ = -1.3 m/s

change in kinetic energy = final kinetic energy  - initial kinetic energy

ΔK =  (¹/₂m₁v₁² + ¹/₂m₂v₂²)  -  (¹/₂m₁u₁² + ¹/₂m₂u₂²)

ΔK =  ( ¹/₂ x 2.5 x -1.3² + ¹/₂ x 5 x 5.2²)  -   (¹/₂ x 2.5 x 3.9² + ¹/₂ x 5 x 2.6²)

ΔK =  69.7125 J - 35.9125 J

ΔK = 33.8 J

aabdulsamir

Answer:

A = 2.5m/s

B = 0.875J

C = 6.57J

Explanation:

M1 = 2.2kg

U1 = 3.9m/s

M2 = 5.0kg

U2 = 2.6m/s

V1 = ?

V2 = 3.3m/s

A. After the collision, for conservation of linear momentum,

M1U1 + M2U2 = M1V1 + M2V2

Solve for V1

V1 = (M1U1 + M2U2 - M2V2) / M1

V1 = (2.5*3.9) + (5.0*2.6) - (5.0*3.3) / 2.5

V1 = (9.75 + 13 - 16.50) / 2.2

V2 = 6.25 / 2.5 = 2.5m/s

b. The change in total kinetic energy after the collision is the difference between the sum of the initial kinetic energy of the two blocks and final kinetic energy of the two blocks

∇K.E = [½(M1U1²) + ½(M2U2²)] - [½M1V1² + ½M2V2²]

∇K.E = [½(2.5 * 3.9²) + ½(5.0*2.6²)] - [½(2.5*2.5²) + ½(5*3.3²)]

∇K.E = (19.01 + 16.90) - (7.81 + 27.225)

∇K.E = 35.91 - 35.035

∇K.E = 0.875J

c.

From the previous equation where we defined V2,

V1 = [(M1U1 + M2U2) - M2V2] / M2

V1 = [(2.5*3.9) * (5.0*2.6) - (5.0*5.2)] / 2.5

V1 = [(9.75 + 13) - 26] / 2.5

V1 = -1.3m/s

∇K.E = ½(M1U1² + M2U2²) - ½(M1V1² + M2V2²

∇K.E = ½(2.5*3.9² + 5.0*2.6²) - ½(2.5* (-1.3)² + 5.0 * 3.3²)

∇K.E = 35.915 - 29.34

∇K.E = 6.57J

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