Question

In: Physics

A 7.50g bullet traveling at 490m/s embeds itself in a 1.59kg wooden block at rest on...

A 7.50g bullet traveling at 490m/s embeds itself in a 1.59kg wooden block at rest on a frictionless surface. The block is attached to a spring with k = 89.0N/m

Find the Period

Find the amplitude of the subsequent simple harmonic motion.

Expert Solution

The period of the block oscillating simple harmonic motion is

Here, Spring constant is k and mass of the block attached to the spring is M.

Subsitute 7.50 g for m and 89.0 N/m for k in the above equation,

Rounding off to three signifiacnt figures, the period of the simple harmonic motion is 0.839 s.

The kinetic energy associated with the moving bullet is

Here, mass of the bullet is m and velocity of the bullet is v.

The elastic potential energy that will be stored by the spring in compressing it is

Here, compression of the spring is x. It is also equal to the amplitude of the simple harmonic motion.

From the law of conservation of energy, the kinetic energyassociated with the moving bullet is converted into the elastic potential energy of the spring. Thus,

This gives the amplitude of the oscillation.

Subsitute 7.50 g for m, 490 m/s for v and 89.0 N/m for k in the above equation,

Rounding off to three significant figures, the mplitude of the simple harmonic motion is 4.49 m.

genius_generous answered 1 month ago

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