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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). The maximum rise of centre of mass of the ball of massm_(2) is

Answer» <html><body><p>`0.20m`<br/>`0.50 m`<br/>`0.75 m`<br/>`1m`</p>Solution :Ball of mass `m_(1)` falls freely <a href="https://interviewquestions.tuteehub.com/tag/till-709848" style="font-weight:bold;" target="_blank" title="Click to know more about TILL">TILL</a> thread becomes taut. At that instant inclination `theta` of the thread with the vertical be given by<br/> `sintheta=(((sqrt(<a href="https://interviewquestions.tuteehub.com/tag/3-301577" style="font-weight:bold;" target="_blank" title="Click to know more about 3">3</a>))/2l))/l=(sqrt(3))/2` or `theta=60^(@)` <br/> Ball of mass `m_(1)` falls freely through height. <br/> <img src="https://d10lpgp6xz60nq.cloudfront.net/physics_images/BMS_VOL2_C01_E01_335_S01.png" width="80%"/> <br/> Velocity of this ball at this instant is `v=sqrt(2gxxl/2)=sqrt(gl)` <br/> It can be resolved into two components. <br/> i. `v cos theta`, along the thread. But thread is inextensible, hence this component decreases to zero (due to tension developed in the thread). <br/> ii. `v sin theta`, perpendicular to the thread. Due to this component ball <a href="https://interviewquestions.tuteehub.com/tag/starts-1224822" style="font-weight:bold;" target="_blank" title="Click to know more about STARTS">STARTS</a> to move along a circle whose centre is at `A`. <br/> According to law of conservation of <a href="https://interviewquestions.tuteehub.com/tag/energy-15288" style="font-weight:bold;" target="_blank" title="Click to know more about ENERGY">ENERGY</a>, <br/> Kinetic energy of ball `m_(1)` just before collision `=` Its kinetic energy at `AC +` Further loss of its potential energy <br/> `1/2m_(1)u_(1)^(2)=1/2m_(1) (sqrt(gl)sintheta)^(2)+m_(1)g(l-lcostheta)` <br/> `u_(1)=4m//s` <br/> According to law of conservation of momentums <br/> `m_(1)v_(1)+m_(2)v_(2)=m_(1)u_(1)+m_(2)u_(2)` <br/> and coefficient of restitution `e=(v_(2)-v_(1))/(u_(2)-u_(1))=1` <br/> 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 <br/> `h=v_(2)^(2)/(2g)=0.20m`</body></html>


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