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Two particles are thrown horizontally in opposite directions from the same point from a height h with velocities 4 ms^(-1) and 3 ms^(-1) . What is the separation between them when their velocities are perpendicular to each other? |
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Answer» Solution :Let us choose a coordinate system with x-axis along horizontal and the y-axis along vertical direction. Then, USING EQUATION `vec v = vec u + vec at`, we GET velocities as `vec v_(1) =4 hat i-gt hat j and vec v_(2)=3 hati -gt hat j`. For velocities to be perpendicular `vec v_(1) cdot vec v_(2) =0` i.e., `(4hat i -g hati j) =0" or "12- g^(2)t^(2)=0` `rarr" "t= sqrt((12)/(g^(2)))=0.35 s` In vertical direction both have same initial velocities and THUS has no separation. Hence, there is only horizontal separation. Therefore, separation `=(v_(1)+v_(2))t =7(0.35) =2.45 m`. |
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