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Two balls of masses m_(1)=100 g and m_(2)=300 g are suspended from point A by two equal inextensible threads, each of length l=32/35m. Ball of mass m_(1) is drawn aside and held at the same level as A but at a distance (sqrt(3)/2)l from A, as shown in Fig. When ball m_(1) is released, it collides elastically with the stationary ball of mass m_(2). Velocity u_(1) with which the hall of mass m_(1) collides with the other ball is |
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Answer» `1m//s` `sintheta=(((sqrt(3))/2l))/l=(sqrt(3))/2` or `theta=60^(@)` Ball of mass `m_(1)` falls freely through height.  Velocity of this ball at this instant is `V=sqrt(2gxxl/2)=sqrt(gl)` It can be resolved into two components. i. `v cos theta`, along the thread. But thread is inextensible, hence this component decreases to zero (due to tension developed in the thread). ii. `v sin theta`, perpendicular to the thread. Due to this component ball starts to move along a circle whose centre is at `A`. According to law of conservation of energy, Kinetic energy of ball `m_(1)` just before collision `=` Its kinetic energy at `AC +` Further loss of its potential energy `1/2m_(1)u_(1)^(2)=1/2m_(1) (sqrt(gl)sintheta)^(2)+m_(1)G(l-lcostheta)` `u_(1)=4m//s` According to law of conservation of momentums `m_(1)v_(1)+m_(2)v_(2)=m_(1)u_(1)+m_(2)u_(2)` and coefficient of restitution `e=(v_(2)-v_(1))/(u_(2)-u_(1))=1` substituting `u_(1)=4 m//s, u_(2)=, m_(1)=0.1 kg` and `m_(2)=0.3kg, v_(2)=2m//s,` the height to which it rises is `h=v_(2)^(2)/(2g)=0.20m` |
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