question_answer

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

A)  \[7.5\text{ }J\]         

B)  \[5\text{ }J\]

C)  \[2.5\text{ }J\]         

D)  \[10\text{ }J\]

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 \[5\text{ }kg,\] and \[{{u}_{2}}\] of\[2.5\text{ }kg\]. Then Momentum before collision = Momentum after collision \[{{m}_{1}}{{u}_{1}}+{{m}_{2}}{{u}_{2}}=({{m}_{1}}+{{m}_{2}}){{v}_{1}}\] Given,  \[{{m}_{1}}=5kg,\,\,\,{{m}_{2}}=2.5k,\,\,\,\,{{u}_{2}}=0\] \[\therefore \] \[5{{u}_{1}}=(5+2.5)v\] \[\Rightarrow \] \[{{u}_{1}}=\frac{7.5}{5}v=1.5v\] ?..(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) \[\Rightarrow \] \[10=7.5{{v}^{2}}\] \[\Rightarrow \] \[{{v}^{2}}=\frac{10}{7.5}=\frac{4}{3}\] Kinetic energy of 1st body after collision \[=\frac{1}{2}{{m}_{1}}{{u}_{1}}^{2}\] \[=\frac{1}{2}\times 5\times {{(1.5v)}^{2}}\] \[=\frac{1}{2}\times 5\times {{(1.5)}^{2}}\times \frac{4}{3}=7.5J\] Note: In an inelastic collision, kinetic energy is not conserved.