A) 9:1
B) 12:1
C) 16 : 1
D) 18 : 1
Correct Answer: A
Solution :
As discussed in question No. 45, ratio of kinetic energy of the body before and after the collision is given by, \[{{\left( \frac{{{u}_{1}}}{{{v}_{2}}} \right)}^{2}}={{\left( \frac{{{m}_{1}}+{{m}_{2}}}{{{m}_{1}}-{{m}_{2}}} \right)}^{2}}\] Given: \[{{m}_{1}}=m\] and \[{{m}_{2}}=2m\] \[\therefore \] \[\frac{K.E{{.}_{before}}}{K.E{{.}_{after}}}={{\left( \frac{m+2m}{m-2m} \right)}^{2}}\] \[=9:1\]You need to login to perform this action.
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