InterviewSolution
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InterviewSolution
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A particle is projected from a point of an angle with the horizontal. At any instant t, if p is the linear momentum and E the kinetic energy, then which of the following graph is/are correct?A. B. C. D. |
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Answer» Correct Answer - a (i) `E_(k)=(P^(2))/(2m)rArrE_(k)propP^(2)` Thus, the graph between `E_(k)` and `P^(2)` will be straight line passing through the origin . Hence, option (d) is correct. (ii) `E_(k)=(1)/(2)mv^(2)=(1)/(2)m(v_(X)^(2)+v_(y)^(2))` `=(1)/(2)m(u cos theta)^(2)+(u sin theta-gt)^(2)` `=(1)/(2)m(u^(2)+g^(2)t^(2)-2ugt sin theta)` ........(i) Thus, the graph between `E_(k)` and t will be parabolic with positiv intercept, Hence, option (b) is correct. (iii) `E_(k)=(1)/(2)mv^(2)=(1)/(2)m(u^(2)-2gy)` `E_(k)=-mgy+(1)/(2)mu^(2)` The graph is a straight line with negaive slope and positive intercept. Hence, option (a) is wrong. (iv) From eqn. (i), `E_(k)=(1)/(2)m[u^(2)+g^(2)t^(2)-2ugt sin theta]` `E_(k)=(1)/(2)m[u^(2)+g((x)/(v_(x)))^(2)-2ug((x)/(v_(x)))sintheta]` Thus, the graph between `E_(k)` and x is a parabolic with positive intercept Hence option (c) is correct. |
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