Question

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.

Answer 1

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.