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 the equilibrium position, it hits another block and comes to rest momentarily, while…

Question

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 the equilibrium position, it hits another block and comes to rest momentarily, while the other block moves with velocity 3 m s - 1 . The total initial energy of the spring is:

Aicharya · Accepted Answer

By mechanical energy conservation between compression positions

x

and

\frac{x}{2}

\frac{1}{2} k x^{2} = \frac{1}{2} k {(\frac{x}{2})}^{2} + \frac{1}{2} m v^{2}

\frac{1}{2} k x^{2} - \frac{1}{2} k \frac{x^{2}}{4} = \frac{1}{2} m v^{2}

\frac{1}{2} k x^{2} (\frac{3}{4}) = \frac{1}{2} m v^{2}

v = \sqrt{\frac{3 k x^{2}}{4 m}} = \sqrt{\frac{3 k}{m}} \frac{x}{2}

On collision with a block at rest

∵

Velocities are exchanged

\Rightarrow

elastic collision between identical masses.

∴ v = 3 = \sqrt{\frac{3 k}{m}} \frac{x}{2}

\Rightarrow 6 = \sqrt{\frac{3 k}{m}} x

\Rightarrow x = 6 \sqrt{\frac{m}{3 k}}

∴

The initial energy of the spring is

U = \frac{1}{2} k x^{2} = \frac{1}{2} k \times 36 \frac{m}{3 k} = 6 m

U = 6 \times 0.1 = 0 .6 J

A block of mass m = 0.1 kg is connected to a spring of unknown spring constant k. It is compressed to a…

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