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NCERT Exemplar · Q37

Q.A simple pendulum has a bob A of mass mm on a string of length 1 m. The bob A is pulled aside until the string is horizontal (level with the pivot) and released from rest. At the lowest point of its swing, A strikes a second bob B of the same mass mm that rests on a table exactly at that lowest point. The collision is elastic and the sizes of the bobs may be neglected. Taking g=9.8 m s−2g = 9.8\ \text{m s}^{-2}, calculate

(a) the height to which bob A rises after the collision, and
(b) the speed with which bob B starts to move.
Rajasthan RbseShort· 3mImportance★★★★★est
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Falling from horizontal, bob A reaches the bottom with speed v=2gL≈4.4 m s−1v=\sqrt{2gL}\approx 4.4\ \text{m s}^{-1}. In an elastic collision between equal masses the velocities are exchanged: A stops (rises to zero height) and B moves off with ≈4.4 m s−1\approx 4.4\ \text{m s}^{-1}.

Speed of A just before impact

A drops through a height equal to the string length L=1 mL=1\ \text{m}. Energy conservation:

12mv2=mgL  ⟹  v=2gL=2×9.8×1=19.6≈4.43 m s−1\tfrac{1}{2}mv^2 = mgL \implies v = \sqrt{2gL} = \sqrt{2\times 9.8 \times 1} = \sqrt{19.6} \approx 4.43\ \text{m s}^{-1}

Elastic collision of equal masses

For a head-on elastic collision between two equal masses, the moving one stops and the struck one moves off with the incoming speed (the velocities are exchanged).

(a) Height A rises …

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