A 0.0486-kg bullet is fired horizontally into a 3.66-kg wooden block attached to

Question

A 0.0486-kg bullet is fired horizontally into a 3.66-kg wooden block attached to one end of a massless, horizontal spring (k = 882 N/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 amplitude of 0.166 m. What is the speed of the bullet?

Explanation / Answer

Mass of bullet M1 =0.0486 kg

Mass of block M2 = 3.66 kg

Spring constant k =882 N/m

Compression in spring amplitude x =0.166 m

Let vo = speed of bullet at impact,

v = common speed of bullet+block after impact

Consider the bullet+the block+the spring as system.

From the time the bullet gets embedded in the block to the time when the spring undergoes maximum

compression, the mechanical energy of the system is conserved.

Therefore 1/2 * (M1+M2)*v^2 = 1/2 kx^2

Or (M1+M2)*v^2 = kx^2

Or v^2 = kx^2/(M1+M2)

Or v = x * sqrt(k/(M1+M2)) = 0.166 * sqrt(882/(0.0486 + 3.66)

= 0.166 * sqrt(882/3.7086)

= 0.166 * sqrt(237.82)

= 2.55 m/s

Momentum of the system is conserved during the impact. Therefore

M1*v0 = (M1+M2)*v

Or v0 = (M1+M2)*v/M1 = (0.0486+3.66)*2.55/0.0486 = 3.7086 * 2.55/0.0486

= 194.58m/s

Ans:194.58 m/s

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