Saved Bookmarks
| 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. |
|