question_answer

A body of mass 4 kg moving with velocity 12 m/s collides with another body of mass 6 kg at rest. If two bodies stick together after collision, then the loss of kinetic energy of system is :

A)  zero                                     

B)  288 J

C)  172.8 J                                 

D)  144 J

Correct Answer: C

Solution :

                 In an inelastic collision, kinetic energy is not conserved but the total energy and momentum remains conserved. Momentum before collision = Momentum after collision \[{{m}_{1}}{{u}_{1}}+{{m}_{2}}{{u}_{2}}={{m}_{1}}{{v}_{1}}+{{m}_{2}}{{v}_{2}}\] \[\Rightarrow \]               \[4\times 12=(4+6)v\] \[\Rightarrow \]               \[v=4.8\,m/s\] Kinetic energy before collision\[=\frac{1}{2}{{m}_{1}}u_{1}^{2}\]                 \[=\frac{1}{2}\times 4\times {{(12)}^{2}}\]                 \[=288\,J\] Kinetic energy after collision                 \[=\frac{1}{2}({{m}_{1}}+{{m}_{2}}){{v}^{2}}\]                 \[=\frac{1}{2}(10){{(4.8)}^{2}}\] \[=115.2\text{ }J\] \[\therefore \] Loss in kinetic energy\[=288\text{ }J-115.2\text{ }J\] \[=172.8J\]