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A ball of sqrt(3)xx10^(-2) kg hits a hard surface at 45^(@) to normal with speed 4sqrt(2) m//s and rebounds with 8//sqrt(3)m//s , at 60^(@) angle. If ball remains in contact for 0.1 sec, what force does it exert ?

Answer» <html><body><p></p>Solution :During rebounce, <a href="https://interviewquestions.tuteehub.com/tag/horizontal-1029056" style="font-weight:bold;" target="_blank" title="Click to know more about HORIZONTAL">HORIZONTAL</a> component of momentum (parallel to surface) does not change but <a href="https://interviewquestions.tuteehub.com/tag/vertical-726181" style="font-weight:bold;" target="_blank" title="Click to know more about VERTICAL">VERTICAL</a> component (normal to surface) changes by,<br/><img src="https://doubtnut-static.s.llnwi.net/static/physics_images/AKS_NEO_CAO_PHY_XI_V01_MP1_C05_SLV_015_S01.png" width="80%"/><br/>`Delta p = mv_(<a href="https://interviewquestions.tuteehub.com/tag/2-283658" style="font-weight:bold;" target="_blank" title="Click to know more about 2">2</a>)cos 60-(-mv_(1)cos <a href="https://interviewquestions.tuteehub.com/tag/45-316951" style="font-weight:bold;" target="_blank" title="Click to know more about 45">45</a>)`<br/>`= m(8//sqrt(3))1//2+m(4sqrt(2))1//sqrt(2)=m((4//sqrt(3))+4)`<br/>`= sqrt(3)xx10^(-2)((4//sqrt(3))+4)=(4+4sqrt(3))xx10^(-2)=10.92xx10^(-2)N-s`<br/>`therefore <a href="https://interviewquestions.tuteehub.com/tag/f-455800" style="font-weight:bold;" target="_blank" title="Click to know more about F">F</a>=(Delta p)/(Delta t)=(10.92xx10^(-2))/(0.1)=10.92xx10^(-1)N`,<br/>Normal to surface.</body></html>


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