A body of mass m strikes another body at rest of mass ( m / 9 ) Assuming the impact to be inelastic the fraction of the initial kinetic energy transformed into heat during the contact is

Question

A body of mass m strikes another body at rest of mass ( 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

Tardigrade Admin · Accepted Answer

The loss in kinetic energy which is transformed into heat
=

\frac{1}{2} (\frac{m _{1} m _{2}}{m _{1} + m _{2}}) (u_{1} - u_{2})^{2}

Here

m_{1} = m, m_{2} = \frac{m}{9}, u_{1} = u_{2} = 0

Loss

△ E = \frac{1}{2} (\frac{m \times \frac{m}{9}}{m + \frac{m}{9}}) \times (u + 0)^{2} = \frac{1}{2} \frac{m}{10} u^{2}

Now, initial kinetic energy =

\frac{1}{2} m u^{2}

Required fraction =

\frac{l oss in kin e t i c e n er g y}{ini t ia l kin e t i c e n er g y}

= \frac{\frac{1}{2} \frac{m}{10} u ^{2}}{\frac{1}{2} m u ^{2}} = \frac{1}{10} = 0.1

Q. A body of mass m strikes another body at rest of mass 9m​ Assuming the impact to be inelastic the fraction of the initial kinetic energy transformed into heat during the contact is

0.1

0.2

0.5

0.64

Q. 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