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An air-track cart with mass m1=0.28kg and initial speed v0=0.75m/s collides with and sticks to a second cart that is at rest initially. if the mass of the second cart is m2=0.43kg, how much kinetic energy is lost as a result of the collision?

Respuesta :

Yna9141
Kinetic energy is calculated through the equation,

   KE = 0.5mv²

At initial conditions,

  m₁:  KE = 0.5(0.28 kg)(0.75 m/s)² = 0.07875 J

  m₂ : KE = 0.5(0.45 kg)(0 m/s)² = 0 J

Due to the momentum balance,

   m₁v₁ + m₂v₂ = (m₁ + m₂)(V)

Substituting the known values,

   (0.29 kg)(0.75 m/s) + (0.43 kg)(0 m/s) = (0.28 kg + 0.43 kg)(V)

   V = 0.2977 m/s

The kinetic energy is,
   KE = (0.5)(0.28 kg + 0.43 kg)(0.2977 m/s)²
   KE = 0.03146 J

The difference between the kinetic energies is 0.0473 J. 
onyebuchinnaji

The kinetic energy lost as a result of the collision is 0.048 J.

The given parameters;

  • mass of the first cart, m₁ = 0.28 kg
  • initial velocity of the cart, u₁ = 0.75 m/s
  • mass of the second cart, m₂ = 0.43 kg
  • initial velocity of second cart, u₂ = 0

The final velocity of both carts after collision is calculated by applying the principle of conservation of linear momentum;

[tex]m_1 u_1 + m_2u_2 = v(m_1 + m_2)\\\\0.28\times 0.75 \ + \ 0.43\times 0= v(0.28 + 0.43)\\\\0.21 = v(0.71)\\\\v = \frac{0.21}{0.71} \\\\v = 0.296 \ m/s[/tex]

The initial kinetic energy before collision is calculated as;

[tex]K.E_i = \frac{1}{2} m_1u_1^2 + \frac{1}{2}m_2u_2^2\\\\K.E_i= \frac{1}{2} \times 0.28 \times 0.75^2 + 0\\\\K.E_i = 0.0788 \ J[/tex]

The final kinetic energy after collision is calculated as;

[tex]K.E_f = \frac{1}{2} (m_1 + m_2)v^2\\\\K.E_f = \frac{1}{2} (0.28 + 0.43)(0.296)^2\\\\K.E_f = 0.0311 \ J[/tex]

The change in kinetic energy is calculated as;

[tex]\Delta K.E = K.E_f - K.E_i\\\\\Delta K.E = 0.0311 \ J \ -\ 0.0788 \ J\\\\\Delta K.E = -0.048 \ J[/tex]

Thus, the kinetic energy lost as a result of the collision is 0.048 J.

Learn more here: https://brainly.com/question/13017090

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