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Physics - Center of Mass Question with Solution | TestHub

PhysicsCenter of MassMomentum conservationMedium3 minQB

PhysicsMediumsingle choice

Passage / Comprehension

As shown in figure there is a small bob having mass 1 kg is attached to the string having negligible mass. Bob is released as shown in figure. Initially when spring is in natural length. (g = 10 m/s $^{2}$ )

If there is no loss in mechanical energy due to collision then find the maximum compression in the spring?

Options:

Answer:

Solution:

Bob's speed at the bottom: $v = 2 gL (1 - cos 6 0^{\circ}) = 2 \times 10 \times 25.6 \times 0.5 = 16 m / s$ . Elastic collision, bob (1 kg) on block (3 kg): block gets $v_{b} = \frac{2 ( 1 ) ( 16 )}{1 + 3} = 8 m / s$ . Compression against friction: $\frac{1}{2} (3) (8^{2}) = \frac{1}{2} k x^{2} + μ m gx \Rightarrow 96 = 100 x^{2} + 4 x$ . $x = \frac{- 4 + 16 + 38400}{200} = 0.96 m = 96 cm$ . Answer: (D)

Stream:JEESubject:PhysicsTopic:Center of MassSubtopic:Momentum conservation

⏱ 3mℹ️ Source: QB

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