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An 1640 kg car stopped at a traffic light is struck from the rear by a 820 kg car. The two cars become entangled, moving along the same path as that of the originally moving car. If the smaller car were moving at 25.6 m/s before the collision, what is the velocity of the entangled cars after the collision?

(a) What is the loss of kinetic energy (Ki - Kf) in the situation described in the example?
kJ

(b) What if the 820 kg car actually moves backwards with a speed of 1.9 m/s right after the collision instead of having a perfectly inelastic collision. What is the velocity of the heavier car immediately after the collision? Use the same convention for positive direction as defined in the example.
m/s

(c) What is the loss of kinetic energy in this case?
kJ

Respuesta :

onyebuchinnaji

(a.i) The velocity of the entangled cars after the collision is 8.53 m/s.

(a.ii) The loss in kinetic energy of the situation described is  179,201.7 J.

(b) The final velocity of the heavier car if the smaller car moved backwards is 11.85 m/s forward.

What is the final velocity of the cars after the collision?

The final velocity of the cars after the collision is calculated by applying the principle of conservation of linear momentum as shown below.

m₁u₁  +  m₂u₂  = v(m₁ + m₂)

where;

  • m₁ is the mass of the smaller car
  • m₂ is the mass of the bigger car
  • u₁ is the initial speed of the smaller car
  • u₂ is the initial speed of the bigger car
  • v is the final speed of the cars

820(25.6) + 1640(0) = v(820 + 1640)

20,992 + 0 = 2460v

v = 20,992 / 2460

v = 8.53 m/s

The lost in kinetic energy of the situation described is calculated as follows;

ΔK.E = K.Ei - K.Ef

K.Ei = ¹/₂m₁u₁²  +   ¹/₂m₂u₂²

K.Ei = ¹/₂(820)(25.6)²  +   ¹/₂(1640)(0)²

K.Ei = 268,697.6 J

K.Ef = ¹/₂(m₁ + m₂)v²

K.Ef = ¹/₂(1640 + 820)(8.53)²

K.Ef = 89,495.91 J

ΔK.E = K.Ei - K.Ef

ΔK.E = 268,697.6 J - 89,495.91 J

ΔK.E =  179,201.7 J

The final velocity of the heavier car if the smaller car moved backwards is calculated as;

m₁u₁  +  m₂u₂  = m₁v₁ + m₂v₂

where;

  • v₁ is the final velocity of the smaller car
  • v₂ is the final velocity of the heavier car

m₁u₁  +  m₂u₂  = m₁v₁ + m₂v₂

820(25.6) + 1640(0) = (820)(-1.9) + 1640(v₂)

20,992 = 1558 + 1640v₂

1640v₂ = 19,434

v₂ = 19,434 / 1640

v₂ = 11.85 m/s

Learn more about loss of kinetic energy here: https://brainly.com/question/7221794

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