JEE Advanced · 2020 · 27 Sep 2020 · Paper 2 · Laws of Motion

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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)

(A) $v_{0}^{2} - 2 g h = \frac{1}{2} g R$ ✓ Correct
(B) $v_{0}^{2} - 2 g h = \frac{3}{2} g R$
(C)the centripetal force required at points x and z is zero
(D)the centripetal force required is maximum at points x and z✓ Correct

Answer: (A), (D)

Solution

By energy conservation between bottom and point y:

\frac{1}{2} m v_{0}^{2} = m g h + \frac{1}{2} m v_{1}^{2}

v_{1}^{2} = v_{0}^{2} - 2 g h

... (i)

At point y, the component of weight along the ramp provides centripetal force:

m g sin 30° = \frac{m v _{1}^{2}}{R}

v_{1}^{2} = \frac{g R}{2}

... (ii)

From (i) and (ii):

v_{0}^{2} - 2 g h = \frac{g R}{2}

✓ (Option A)

At points x and z, the height is less than h, so velocity is greater than at y. Therefore, required centripetal force

\frac{m v ^{2}}{R}

is maximum at these points. ✓ (Option D)

JEE Advanced · 2020 · 27 Sep 2020 · Paper 2 · One or more correct options

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JEE Advanced · 2020 · 27 Sep 2020 · Paper 2 · Laws of Motion

Solution

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