A) Zero
B) \[({{m}_{1}}+{{m}_{2}})g{{t}_{0}}\]
C) \[2({{m}_{1}}+{{m}_{2}})g{{t}_{0}}\]
D) \[\frac{1}{2}({{m}_{1}}+{{m}_{2}})g{{t}_{0}}\]
Correct Answer: C
Solution :
The momentum of the two-particle system, at t = 0 is \[{{\vec{P}}_{i}}={{m}_{1}}{{\vec{v}}_{1}}+{{m}_{2}}{{\vec{v}}_{2}}\] Collision between the two does not affect the total momentum of the system. A constant external force \[({{m}_{1}}+{{m}_{2}})g\] acts on the system. The impulse given by this force, in time t = 0 to \[t=2{{t}_{0}}\] is \[({{m}_{1}}+{{m}_{2}})g\times 2{{t}_{0}}\] \ |Change in momentum in this interval \[=\,|{{m}_{1}}\vec{v}{{'}_{1}}+{{m}_{2}}\vec{v}{{'}_{2}}-({{m}_{1}}{{\vec{v}}_{1}}+{{m}_{2}}{{\vec{v}}_{2}})|\,=2({{m}_{1}}+{{m}_{2}})g{{t}_{0}}\]You need to login to perform this action.
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