Question

A body of mass m strikes another body at rest of mass $ \frac{ m }{ 9 }$ Assuming the impact to be inelastic the fraction of the initial kinetic energy transformed into heat during the contact is

• A. 0.1
• B. 0.2
• C. 0.5
• D. 0.64

Answer 1

Answer: (A)

The loss in kinetic energy which is transformed into heat
= $ \frac{1}{2} \bigg( \frac{ m_1 m_2 }{ m_1 + m_2 } \bigg) ( u_1 - u_2 )^2 $
Here $ m_1 = m, \, m_2 = \frac{ m }{ 9 } , \, u_1 = u_2 = 0 $
Loss $ \triangle E = \frac{1}{2} \Bigg( \frac{ m \times \frac{ m }{ 9 }}{ m + \frac{ m }{ 9 }} \Bigg ) \times ( u + 0 )^2 = \frac{ 1}{ 2} \frac{ m }{ 10 } u^2 $
Now, initial kinetic energy = $ \frac{ 1}{ 2} mu^2 $
Required fraction = $ \frac{ loss \, in \, kinetic \, energy }{ initial \, kinetic \, energy } $
$ = \frac{ \frac{ 1}{ 2} \frac{ m }{ 10 } u^2 }{ \frac{ 1}{ 2} mu^2 } = \frac{ 1}{ 10} = 0.1 $