question_answer

A body of mass 5 kg makes an elastic collision with another body at rest and continues to move in the original direction after collision with a velocity equal to \[\frac{1}{10}th\] of its original velocity. Then the mass of the second body is

A)  4.09 kg          

B)  0.5 kg

C)  5 kg            

D)  5.09 kg  

Correct Answer: A

Solution :

Mass of the first body \[{{m}_{1}}=5kg,\], for elastic collision\[e=1\]. Suppose initially body \[{{m}_{1}}\] moves with velocity  after collision velocity becomes \[\left( \frac{u}{10} \right)\] . Let after collision velocity of M block becomes \[({{v}_{2}})\]. By conservation of momentum \[{{m}_{1}}{{u}_{1}}+{{m}_{2}}{{u}_{2}}={{m}_{1}}{{v}_{1}}+{{m}_{2}}{{v}_{2}}\] or \[5u+M\times 0=5\times \frac{u}{10}+M{{v}_{2}}\] or \[5u=\frac{u}{2}+M{{v}_{2}}\]            ?..(i) Since,        \[{{v}_{1}}-{{v}_{2}}=-e({{u}_{1}}-{{u}_{2}})\] \[\frac{u}{10}-{{v}_{2}}=-1(u)\] or \[\frac{u}{10}+u={{v}_{2}}\] \[\frac{11u}{10}={{v}_{2}}\]                ??(ii) Substituting value of v^ in Eq. (i) from Eq. (ii),      we get                 \[5u=\frac{u}{2}+M\left( \frac{11u}{10} \right)\] or \[5-\frac{1}{2}=M\left( \frac{11}{10} \right)\] or \[M=\frac{9\times 10}{2\times 11}\] or \[M=\frac{45}{11}=4.09kg\]