A 1.00 (10^-2 kg) bullet is fired horizontally into a 2.50 kg wooden block attac

Question

A 1.00 (10^-2 kg) bullet is fired horizontally into a 2.50 kg wooden block attached to one end of a massless horizontal spring (k=845 M/m). The other end of the spring is fixed in place, and the spring is unstrained initially. The block rests on a horizontal, frictionless surface. The bullet strikes the block perpendicularly and quickly comes to a halt within it. As a result of this completely inelastic collision, the spring is compressed along its axis and causes the block/bullet to oscillate with an ampliude of 0.200m.

a) What is the speed of the bullet?

b) What happens to the amplitude of vibration if we double the mass of the bullet? (keeping all other variables the same)

c) What happens to the amplitude of vibration if we double the velocity of the bullet? (keeping all other variables the same)

d) What happens to the amplitude of vibration if we double the mass of the block? (keeping all other variables the same)

Explanation / Answer

a) At the extreme end (when the spring is compressed), the kinetic energy of bullet is completely converted to potential enrgy of spring. At this location compression x = 0.5 * amplitude of oscillation = 0.100m

So, 0.5 *Mbullet * (Vbullte ^2) = 0.5 * k * x^2

Vbullet = sqrt (k * x^2 /Mbullet) = sqrt (845 * (0.1 *0.1)/0.01) = sqrt (845) = 29.06888 m/s

So the speed of bullet was 29.06888m/s

b) From (a) we see Mass is proportional to square of amplitude. So if mass is doubled, amplitude of vibration goes up by factor sqrt(2)

c) Similarly V is proportional to X. So if velocity is doubled, so will the amplitude get doubled

d) Mass of block has no impact on the amplitude of vibration. (It will only have impact on the velocity of spring mass at equilibrium)

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