JEE Advanced 2015 · Paper 1 · Physics Q4

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PhysicsInteger+4Question 4 of 60

Two identical uniform discs roll without slipping on two different surfaces $AB$ and $CD$ (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 $v_2$ in m/s is ( $g=10\ \mathrm{m/s^2}$ )

Your answer

Answer and worked solution

Official answer: 7

Rolling energy. A rolling disc has $KE=\frac12mv^2+\frac12\cdot\frac12mr^2\frac{v^2}{r^2}=\frac34mv^2$ .

Energy equations. Disc 1 drops 30 m and disc 2 drops 27 m, and both end with the same speed $v_f$ :
$\tfrac34v_f^2=\tfrac34(3)^2+10(30)=\tfrac34v_2^2+10(27).$

Solve. $\frac34v_2^2=6.75+30=36.75$ , so $v_2^2=49$ and $v_2=7$ m/s.

Answer: 7

Topics:Rolling motion Conservation of mechanical energy

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JEE Advanced 2015 · Paper 1, Physics question 4