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?

Options

A. (a)
B. (b)
C. (c)
D. (d)

Answer

A, B, C, D

Explanation

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.