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Physics · Horizontal Circular Motion

JEE Advanced 2020 — Paper 2 — Question 6

A student skates up a ramp that makes an angle 30∘30^{\circ} with the horizontal. He/she starts (as shown in the figure) at the bottom of the ramp with speed v0\mathrm{v}_{0} and wants to turn around over a semicircular path xyz of radius R during which he/she reaches a maximum height hh (at point yy ) from the ground as shown in the figure. Assume that the energy loss is negligible and the force required for this turn at the highest point is provided by his/her weight only. Then ( g is the acceleration due to gravity)

Question figure
  1. Option A:

    v02−2gh=12g g\mathrm{v}_{0}^{2}-2 \mathrm{gh}=\frac{1}{2} g \mathrm{~g}

    Correct
  2. Option B:

    v02−2gh=32gR\mathrm{v}_{0}^{2}-2 \mathrm{gh}=\frac{\sqrt{3}}{2} \mathrm{gR}

  3. Option C:

    the centripetal force required at points x and z is zero

  4. Option D:

    the centripetal force required is maximum at points x and z

    Correct

Answer: A, D

Step-by-step solution

Let v be the speed at y

Energy conservation ⇒12mv02=mgh+12mv2\Rightarrow \frac{1}{2} \mathrm{mv}_{0}^{2}=\mathrm{mgh}+\frac{1}{2} m v^{2}

FBD at y⇒mgsin⁡30∘=mv2Ry \Rightarrow m g \sin 30^{\circ}=\frac{m v^{2}}{R}

From (1) and (2)

⇒v02−2gh=gR2\Rightarrow \mathrm{v}_{0}^{2}-2 \mathrm{gh}=\frac{\mathrm{gR}}{2}, (A) option is correct.

(D) is true as on the circular path he/she will have maximum speed at x and z .

Answer key and solution verified before publishing.

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Exam
JEE Advanced 2020
Paper
Paper 2
Subject
Physics
Chapter
Horizontal Circular Motion
Topic
Dynamics of circular motion
A student skates up a ramp that makes an angle 30 ° with the… | JEE Advanced 2020 PYQ with Solution · DhiX AI