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A ball of mass 50 gm is dropped from a height h=10m. It rebounds losing 75 percent of its kinetic energy. If it remains in contanct with the ground for `Deltat=0.01` sec. the impulse of the impact force is (take `g=10 m//s^(2)`)A. 1.3 N-sB. 1.06 N-sC. 1300 N-sD. 1.05 |
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Answer» Correct Answer - B `v_(1)=sqrt(2gh)=sqrt(2xx10xx10)=10sqrt(2)` `k_(2)=1/4k_(1)rArr v_(2)^(2)=1/4v_(1)^(2)` `:. v_(2)=(v_(1))/2=5sqrt(2)` `|DeltaP|=|-mv_(2)-(mv_(1))|=m|-v_(2)-v_(1)|` `|DeltaP|=50xx10^(-3)xx3/2xx10sqrt(2)=(15xx10^(-1))/(sqrt(2))` `J=DeltaP=1.05 N-s` |
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