A particle of mass m1 moving with u1 velocity collide with another particle of mass m2 which is initially in rest. If collision is perfectly inelastic then the fractional loss in the kinetic energy is 

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  1. \(\dfrac{1}{4} m_1 u_1^2\)
  2. \(\dfrac{m_2}{m_1}\)
  3. \(\dfrac{m_2}{m_1 + m_2}\)
  4. \(\dfrac{m_1}{m_1+m_2}\)

Answer (Detailed Solution Below)

Option 3 : \(\dfrac{m_2}{m_1 + m_2}\)
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Detailed Solution

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Concept:

  • perfectly elastic collision is defined as one in which there is no loss of kinetic energy in the collision.
  • An inelastic collision is one in which part of the kinetic energy is changed to some other form of energy in the collision.

Explanation:

F1 Prabhu.Y 18-09-20 savita D4

  • By the law of conservation of momentum

⇒ m1u1 = (m1 + m2)V

\(\Rightarrow V = \frac{m_1u_1}{m_1+m_2}\)  

  • By the law of conservation of kinetic energy

\(\Rightarrow \frac{1}{2}m_1u^2_1=\frac{1}{2}(m_1+m_2)V^2\)

  • The ratio of kinetic energy before and after the collision is 

\(\frac{KE_f}{KE_i}=\frac{\frac{1}{2}(m_1+m_2)V^2}{\frac{1}{2}mu^2_1}=\frac{\frac{1}{2}(m_1+m_2)[\frac{m_1u_1}{m_1+m_2}]^2}{\frac{1}{2}mu^2_1}=\frac{m_1}{m_1+m_2}\)

  • The fractional loss in the kinetic energy is

\(\Rightarrow \frac{KE_i-KE_f}{KE_i}=\frac{KE_i([1-\frac{KE_f}{KE_i}])}{KE_i}=\frac{KE_i([1-\frac{m_1}{m_1+m_2}])}{KE_i}=\frac{m_2}{m_1+m_2}\)

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