A) 4
B) 6
C) 8
D) 10
Correct Answer: A
Solution :
The velocity of B just after collision with A is \[{{v}_{B}}=\frac{({{m}_{B}}-{{m}_{A}}){{u}_{B}}}{{{m}_{B}}+{{m}_{A}}}+\frac{2{{m}_{A}}{{u}_{A}}}{{{m}_{A}}+{{m}_{B}}}\] \[=\frac{0+2m\times 9}{m+2m}=6m/s\] The collision between B and C is completely inelastic. \[\therefore \,\,\,{{m}_{B}}{{v}_{B}}=({{m}_{B}}+{{m}_{C}})v\] \[\therefore \,\,\,v=\frac{6\times 2m}{2m+m}=4m/s.\]You need to login to perform this action.
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