InterviewSolution
Saved Bookmarks
InterviewSolution
| 1. |
A particle is projected from a point at an angle with the horizontal at `t=0`. At an instant t, if p is linear momentum, x is horizontal displacement, y is vertical displacement, and E is kinetic energy of the particle, then which of the following graphs are correct?A. (a) B. (b) C. (c) D. (d) |
|
Answer» Correct Answer - A::B::C::D At any time t, `v=sqrt(u^2+g^2t^2-2ug tsintheta)` `E=1/2mv^2=1/2m(u^2+g^2t^2-2ug t sin theta)` Hence, `E-T` graph is parabolic `E=(p^2)/(2m)` Hence, `E-p^2` graph is straight line through origin `E=1/2m u^2-mgy` Putting `y=xtantheta-(gx^2)/(2u^2cos^2theta)` we get `E=1/2m u^2-mg x tan theta+(mx^2x^2)/(2u^2cos^2theta)` Hence, `E-y` graph is a straight line and `E-x` graph is parabolic. |
|