A block of mass M = 5.70 kg, at rest on a horizontal frictionless table, is atta

Question

A block of mass M = 5.70 kg, at rest on a horizontal frictionless table, is attached to a rigid support by a spring of constant k = 5570 N/m. A bullet of mass m = 9.20 g and velocity ModifyingAbove v With right-arrow of magnitude 520 m/s strikes and is embedded in the block (the figure). Assuming the compression of the spring is negligible until the bullet is embedded, determine (a) the speed of the block immediately after the collision and (b) the amplitude of the resulting simple harmonic motion.

Explanation / Answer

Just after impact, the momentum of block plus bullet must equal the original momentum of the bullet.
bullet momentum = 520 x 0.0092 = 4.784
the combined mass of block and bullet M = 5.7 +.0092 = 5.7091 kg
So the momentum M.v = 4.784 so v = 0.8379 m/s

The initial velocity of the block plus bullet is 0.8379 m/s.

Now it is a matter of calculating how far the spring is compressed before the block stops moving. This distance will be the amplitude of the oscillations. When the block stops, all of its kinetic energy has been transferred to the spring as potential energy.

Initial KE of the block+bullet = 0.5 x 5.7091 x (0.8379)^2 = 2.00411J

The PE stored in a spring = 0.5 k.x^2 where k is the spring constant and x is the displacement of the spring.
The KE and PE must equal, so 2.00411 = 0.5 k.x^2
the spring displacement and the amplitude x = SQRT( 2.00411/(0.5k) )
= 0.0268

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