Two particles start moving in the XY-plane from M to N. A particle detector is kept at the location C. The detector gives a signal if a particle touches the X-axis and detects the distance of the particle from the location C. The particle A follows a downward facing parabolic path represented by −x^2+bx+c with a vertex P, where b,c are integers. The second particle B moves along a linear path.
The position of the detector is such that the square of the magnitude of CM is 2 times of the square of the perpendicular distance of the point C from the line N.
If the minimum distance from the point C along the X-axis where particle A can be detected by the detector is k unit, then find the value of k
Answers
Correct option is
A
1 N along y-axis
Since the graph between x and t the is a straight line and passing through origin.
∴ x=t
Since the graph between y and t is a parabola,
∴y=t
2
∴v
x
=
dt
dx
=1 and a
x
=
dt
dv
x
=0
and v
y
=
dt
dy
=2t and a
y
=2ms
−2
The force acting on the particle is
F=ma
y
=(0.5kg)(2ms
−2
)= 1 N along y-axis
If the minimum distance from the point C along the X-axis where particle A can be detected by the detector is k unit, then find the value of k
1 N along y-axis
Since the graph between x and t the is a straight line and passing through origin.
∴ x=t
Since the graph between y and t is a parabola,
∴y=t 2
∴v x
= dt dx
=1 and a x
= dt dv x
=0 and v y
= dt dy
=2t and a y
=2ms −2
The force acting on the particle is
F=ma y =(0.5kg)(2ms −2 )
= 1 N along y-axis