A particle slides down a frictionless parabolic (y + x2 ) track (A – B – C) starting from rest at point A (Fig. 4.2). Point B is at the vertex of parabola and point C is at a height less than that of point A. After C, the particle moves freely in air as a projectile. If the particle reaches highest point at P, then
(a) KE at P = KE at B
(b) Height at P = height at A
(c) Total energy at P = total energy at A
(d) Time of travel from A to B = time of travel from B to P.

A particle slides down a frictionless parabolic (y + x2 ) track (A – B – C) starting from rest at point A (Fig. 4.2). Point B is at the vertex of parabola and point C is at a height less than that of point A. After C, the particle moves freely in air as a projectile. If the particle reaches highest point at P, then
(a) KE at P = KE at B
(b) Height at P = height at A
(c) Total energy at P = total energy at A
(d) Time of travel from A to B = time of travel from B to P.

This is a Multiple Choice Questions as classified in NCERT Exemplar
Answer- c
Explanation– as the given track y=x2 is a frictionless track thus total energy will be same throughout the journey.
Hence total energy at A = total energy at P . at B the particle is having only Ke but at P some KE is convert
Similar Questions for you
Please find the solution below:
after 10 kicks,
v? = 3tî v? = 24cos 60°î + 24sin 60°? = 12î + 12√3?
v? = v? – v? = (12 – 3t)î + 12√3?
It is minimum when 12 - 3t = 0 ⇒ t = 4sec
ω = θ² + 2θ
α = (ωdω)/dθ = (θ² + 2θ) (2θ + 2)
At θ = 1rad.
ω = 3rad/s and α = 12rad/s²
a? = αR = 12 m/s² a? = ω²R = 9 m/s² A? = √ (a? ² + a? ²) = 15 m/s²
a? = v? ²/4r
a_A? = (v? ²/r²) × r = v? ²/r
a_A = 3v? ²/4r
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Physics NCERT Exemplar Solutions Class 11th Chapter Four 2025
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