JEE Advanced · 2015 · 24 May 2015 · Paper 1 · Work, Power & Energy

Two identical uniform discs roll without slipping on two different surfaces

A B

and

C D

(see figure) starting at

A

and

C

with linear speeds

v_{1}

and

v_{2}

, respectively, and always remain in contact with the surfaces. If they reach

B

and

D

with the same linear speed and

v_{1} = 3 m / s

, then

v_{2}

m / s

(g = 10 m / s^{2})

Answer: 7

Solution

For a disc rolling without slipping, total kinetic energy is

\frac{3}{4} m v^{2}

. Using energy conservation for disc 1:

\frac{3}{4} m (3)^{2} + m g (30) = \frac{3}{4} m v_{f}^{2}

. For disc 2:

\frac{3}{4} m v_{2}^{2} + m g (27) = \frac{3}{4} m v_{f}^{2}

. Since both reach the same final speed, equating:

\frac{3}{4} \cdot 9 + 300 = \frac{3}{4} v_{2}^{2} + 270

. Solving:

\frac{27}{4} + 30 = \frac{3}{4} v_{2}^{2}

, giving

v_{2}^{2} = 49

, hence

v_{2} = 7

m/s.

JEE Advanced · 2015 · 24 May 2015 · Paper 1 · Integer answer