question_answer

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

A)  zero         

B)  \[288\text{ }J\]

C)  \[172.8\text{ }J\]       

D)  \[144\text{ }J\]

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 \[{{m}_{1}}{{u}_{1}}+{{m}_{1}}{{u}_{2}}={{m}_{1}}{{v}_{1}}+{{m}_{2}}{{v}_{2}}\] \[\Rightarrow \] \[4\times 12=(4+6)v\] \[\Rightarrow \] \[v=4.8m/s\] Kinetic energy before collision \[=\frac{1}{2}{{m}_{1}}{{u}_{1}}^{2}\] \[=\frac{1}{2}\times 4\times {{(12)}^{2}}\] \[=288J\] Kinetic energy after collision \[=\frac{1}{2}({{m}_{1}}+{{m}_{2}}){{v}^{2}}\] \[=\frac{1}{2}(10){{(4.8)}^{2}}\] \[=115.2J\] \[\therefore \] Loss in kinetic energy \[=288J-115.2J\] \[=172.8J\]