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A ball of sqrt(3) xx 10^(-2) kg hits a hard surface at 45^(@)to normal with speed 4sqrt(2)m/s and rebounds with8//sqrt(3) m//s ' s, at 60^(@) angle. If ball remains in contact for 0.1 sec, what force does it exert ? |
Answer» Solution : During rebounce, horizontal component of momentum (parallel to SURFACE) does not change but vertical component (normal to surface) changes by,  `Deltap = mv_(2) cos THETA-(-mv_(1)cos 45^(@))` `=m(8//sqrt(3))1//2 + m(4sqrt(2))1//sqrt(2) = m((4//sqrt(3))+4)` `=sqrt(3) XX 10^(2)((4sqrt(3)) + 4) = (4+4sqrt(3)) xx 10^(-2) = 10.92 xx 10^(-2) N -s` `therefore F =(Deltap)/(Deltat) = (10.92 xx 10^(-2))/0.1 = 10.92 xx 10, N`, normal to surface. |
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