question_answer

A body of mass 5 kg strikes another body of mass 2.5 kg initially at rest. The bodies after collision coalesce and begin to move as whole with a kinetic energy of 5 J. The kinetic energy of the first body before collision is :

A)  7.5 J                                     

B)  5 J

C)  2.5 J                                     

D)  10 J

Correct Answer: A

Solution :

                This is a case of inelastic collision, in which momentum is conserved, and total energy. Let\[{{u}_{1}}\]be velocity of body of mass 5 kg, and\[{{u}_{2}}\]of 2.5 kg. Then Momentum before collision = Momentum after collision                 \[{{m}_{1}}{{u}_{1}}+{{m}_{2}}{{u}_{2}}=({{m}_{1}}+{{m}_{2}}){{v}_{1}}\] Given, \[{{m}_{1}}=5\,kg,\,{{m}_{2}}=2.5\,kg,\,{{u}_{2}}=0\] \[\therefore \]  \[5{{u}_{1}}=(5+2.5)v\] \[\Rightarrow \]               \[{{u}_{1}}=\frac{7.5}{5}v=1.5\,v\]                   ...(i) Also, kinetic energy after collision                 \[=\frac{1}{2}({{m}_{1}}+{{m}_{2}}){{v}^{2}}\] Given,    \[5=\frac{1}{2}({{m}_{1}}+{{m}_{2}}){{v}^{2}}\]                            ... (ii) \[\Rightarrow \]               \[10=7.5\text{ }{{v}^{2}}\] \[\Rightarrow \]               \[{{v}^{2}}=\frac{10}{7.5}=\frac{4}{3}\] Kinetic energy of 1st body after collision                 \[=\frac{1}{2}{{m}_{1}}u_{1}^{2}\]                 \[=\frac{1}{2}\times 5\times {{(1.5v)}^{2}}\]                 \[=\frac{1}{2}\times 5\times {{(1.5)}^{2}}\times \frac{4}{3}=7.5\,J\] Note: In an inelastic collision, kinetic energy is not conserved.