A block of mass m = 0.1 kg is connected to a spring of unknown spring constant k. It is compressed to a distance x from its equilibrium position and released from rest. After approaching half the distance (x/2) from equilibrium position, it hits another block and comes to rest momentarily, while the other block moves with a velocity 3 ms ⁻¹. The total initial energy of the spring is: (a) 1.5 J (b) 0.6 J (c) 0.3 J (d) 0.8 J

Answer: (b). The block stops and the other block moves at 3 m/s, so (equal-mass elastic collision) the block's speed at x/2 was 3 m/s. KE at x/2: 1/2(0.1)(3²) = 0.45 J This equals spring PE released: 1/2k(x² - x²/4) = 3/4E₀ E₀ = 4/3(0.45) = 0.6 J So the answer is (b) 0.6 J.

A block of mass m = 0.1 kg is connected to a spring of unknown spring… | JEE Main 2015 (10 Apr, Shift 1)

Q21JEE Main 2015 · 10 APR · SHIFT 1Medium

Spring & Elastic Energy

A block of mass m = 0.1 kg is connected to a spring of unknown spring constant k. It is compressed to a distance x from its equilibrium position and released from rest.

After approaching half the distance $\left(\frac{x}{2}\right)$ from equilibrium position, it hits another block and comes to rest momentarily, while the other block moves with a velocity 3 ms ${}^{-1}$ . The total initial energy of the spring is:

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Answer: (b)

The block stops and the other block moves at $3$ m/s, so (equal-mass elastic collision) the block's speed at $x/2$ was $3$ m/s.
KE at $x/2$ : $\frac{1}{2}(0.1)(3^2) = 0.45$ J
This equals spring PE released: $\frac{1}{2}k\left(x^2 - \frac{x^2}{4}\right) = \frac{3}{4}E_0$
$E_0 = \frac{4}{3}(0.45) = 0.6$ J
So the answer is (b) $0.6$ J.

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