A 3.00 kg block is suspended from a spring with k = 540 N/m. A 48.0 g bullet is

Question

A 3.00 kg block is suspended from a spring with k = 540 N/m. A 48.0 g bullet is fired straight up into the block from directly below with a speed of 122 m/s and becomes embedded in the block. Find the amplitude of the resulting simple harmonic motion. What percentage of the original kinetic energy of the bullet is transfered to mechanical energy of the oscillator? The block is initially at the equilibrium point. How does the upward spring force compare with the downward gravitational force? Once the bullet is embedded, how much lower is the new equilibrium point? Do you recall how to find the speed just after a completely inelastic collision? Can you find the kinetic energy just then? How high is the bullet + block when they stop, with the mechanical energy in the two types of potential energy? You need to find how high this stopping point is above the new equilibrium point.

Explanation / Answer

the original energy of the bullet is

Eo = 1/2 m v0^2 = 1/2 (0.048)( 122)^2 = 357.216 J

bullet block system is

E = 1/2 ( m+ M) ( mv0/m+M)^2

= 1/2 ( 0.048+3)( 0.048(122)/ 0.048+ 3)^2

=5.62 J

E/E0 = 5.62 J/357.216 J = 0.0157 or 1.57 %

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