A) 2m/sec in original direction
B) 2 m/sec opposite to the original direction
C) 4 m/sec opposite to the original direction
D) 4 m/sec in original direction
Correct Answer: A
Solution :
\[{{v}_{1}}=\left( \frac{{{m}_{1}}-{{m}_{2}}}{{{m}_{1}}+{{m}_{2}}} \right)\,{{u}_{1}}+\frac{2{{m}_{2}}{{u}_{2}}}{{{m}_{1}}+{{m}_{2}}}\] Substituting m1 = 0, \[{{v}_{1}}=-{{u}_{1}}+2{{u}_{2}}\] Þ \[{{v}_{1}}=-\,6+2(4)\] \[=2m/s\] i.e. the lighter particle will move in original direction with the speed of 2 m/s.You need to login to perform this action.
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