JEE Advanced 2020 · Paper 2 · Physics Q8

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PhysicsMultiple correct+4 / -2Question 8 of 54

A student skates up a ramp that makes an angle 30° with the horizontal. He/she starts (as shown in the figure) at the bottom of the ramp with speed $v_0$ and wants to turn around over a semicircular path $xyz$ of radius $R$ during which he/she reaches a maximum height $h$ (at point $y$ ) 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)

An inclined plane at 30° to the horizontal. The skater moves straight up the slope, then turns around on the ramp along a semicircle x-y-z of radius R whose highest point y is at height h above the ground; x and z lie on the same level, lower down the slope than y.

v_0^2 - 2gh = \dfrac12 gR

v_0^2 - 2gh = \dfrac{\sqrt3}{2} gR

the centripetal force required at points

x

and

z

is zero

the centripetal force required is maximum at points

x

and

z

Answer and worked solution

Official answer: A, D

(A)/(B) At the top $y$ . The skater moves across the slope, and only the component of weight along the incline, $mg\sin30^\circ=\frac{mg}2$ , points towards the centre of the turn:
$\frac{mv_y^2}R=\frac{mg}2\implies v_y^2=\frac{gR}2.$
Energy conservation from the bottom:
$v_0^2=v_y^2+2gh\implies v_0^2-2gh=\tfrac12gR.$
(A) is true and (B) is false.

(C)/(D) At $x$ and $z$ . These are the lowest points of the semicircle, where the speed is greatest. The centripetal force $mv^2/R$ is therefore largest there, not zero. (D) is true and (C) is false.

(A), (D)

Topics:Circular motion dynamics Conservation of mechanical energy

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JEE Advanced 2020 · Paper 2, Physics question 8