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A man of mass m climbs a rope of length L suspended below a balloon of mass M. The balloon is stationary with respect to ground. If the man begins to climb up the rope at a speed v_("rel"). (relative to rope) in what direction and with what speed (relative to gound) with the balloon move ? |
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Answer» <P> Solution :The change in momentum `Delta p` of the ball during `Delta = 0.01` sec is known as the impulse `(F Delta t)` of the force of impact`rArr F Delta t = Delta p` `|Delta VEC(p)|=|m vec(v)_(2)-m vec(v)_(1)|= m (v_(1)+v_(2))` `because m vec(v)_(1)& m vec(v)_(2)` are ANTIPARALLEL `rArr F delta t = m (v_(1)+v_(2)) ""`...(1)  Since, the ball falls through a HEIGHT h, and its velocity just before STRIKING the surface is `sqrt(2gh)`. As the ball losses 75percent of its total mechanical energy that is knetic energy. `rArr 1//2 mv_(2)^(2)=1//2 m v_(1)^(2)(1-75//100)` `rArr v_(2)=(v_(1))/(2)=(sqrt(2gh))/(2)=sqrt((gh)/(2))` Substitution of the values of `v_(1)` and `v_(2)` in (1) yields `F Delta t = 50xx10^(-3)xx3 sqrt((gh)/(2))=50xx10^(-3)xx 3 sqrt(((9.8)(10))/(2))=1.05` N - sec. |
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