Three identical particles A, B and C (each$$ of mass $$m) lie on a smooth horizontal table. Light inextensible strings which are just taut connect AB and BC and ABC is $$1350$$ as shown in diagram. An impulse J is applied to the particle C in the direction BC for a very short time interval. Then just after applying impulse J,The speed of the particle ‘A’ is $$nJ7m$$. The value of n is

Question

Three identical particles A, B and C (each$$ of mass $$m) lie on a smooth horizontal table. Light inextensible strings which are just taut connect AB and BC and ABC is $$1350$$ as shown in diagram. An impulse J is applied to the particle C in the direction BC for a very short time interval. Then just after applying impulse J,The speed of the particle ‘A’ is $$nJ7m$$. The value of n is

Infinity Learn · Accepted Answer

Impulse $$=$$ΔP Along $$X-axis$$: $$J2cos45=$$ $$mv1+mv1$$  $$J22=2mv1--(1)Along$$ $$Y-axis$$: $$J2sin45=$$ $$mv2$$ $$J22=mv2--(2)(1)=(2)$$ $$v2=2v1---(3)$$ $$v3=v1+v22J-J2=mv3$$ $$J-22mv1=mv1+v22$$ $$J=7mv12$$ $$v1=vA=J27m----(4)$$ $$v2=2v1=2J27mfrom$$ $$(3)---(5)vB=v12+v22=5v12=10J7mvC=v1+v22=3v12=32J27m=3J7mv1y=22J7m$$;$$J1=3Jmand$$  $$J2=2J7II$$ method    The external impulse applied to C causes both strings to jerk exerting internal impulses $$J1$$ and $$J2V2=v1x$$−−−−−−−$$(1)J1cos$$⁡$$45$$∘−$$J2=mv1x$$−−−−−−−−−$$(3)J1cos$$⁡$$45$$∘$$=mv1y$$−−−−−−−$$(4)J$$−$$J1=mv$$−….....−$$(5)Also$$ velocities of B&C along BC are are equal i.e.,  $$v1ycos450+v1xcos450=v$$−−−−−$$6After$$ solving we get,  $$v2=v1x=2J7mv=3J7m$$;$$v1y=22J7m$$;$$VA=2J7m$$,$$VB=10J7m$$,$$J1=3Jmand$$  $$J2=2J7$$

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