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 ?

A ball of sqrt(3)xx10^(-2) kg hits a hard surface at 45^(@) to normal

1.	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» Solution :During rebounce, HORIZONTAL component of momentum (parallel to surface) does not change but VERTICAL component (normal to surface) changes by, `Delta p = mv_(2)cos 60-(-mv_(1)cos 45)` `= m(8//sqrt(3))1//2+m(4sqrt(2))1//sqrt(2)=m((4//sqrt(3))+4)` `= sqrt(3)xx10^(-2)((4//sqrt(3))+4)=(4+4sqrt(3))xx10^(-2)=10.92xx10^(-2)N-s` `therefore F=(Delta p)/(Delta t)=(10.92xx10^(-2))/(0.1)=10.92xx10^(-1)N`, Normal to surface.