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A rifle bullet with mass 8.00 g and an initial horizontal velocity strikes and embeds itself in a block with mass 0.992 kg that rests on a frictionless surface and is attached to one end of an ideal spring. The other end of the spring is attached to the wall. The impact compresses the spring a maximum distance of 18.0 cm. 
After the impact, the block moves in SHM with a period of 0.314 s. What was the initial speed of the bullet?
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Added Wed, 10 Jun '15
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What are our givens?

m_bul = 0.008 kg
m = 0.992 kg
x = 0.18 m
T = 0.314 s

To find the spring constant of the spring we can use its period and its mass.

Find the Angular Speed of the Block
We know omega = 2Pi / T
Thus
Omega = 2 * Pi / 0.314 = 2.001 rad / sec


Find the Spring Constant of the Spring. 
We know omega = sqrt(k / m) for a spring.
Solving for K we find,
K = omega^2 * m
Thus
K = (2.001 rad/sec)^2 * 0.992 kg = 3.972 N/m

Find the speed of the Bullet Block system after Impact
We can use the law of conservation of energy to find the speed of the Bullet Block system's speed after impact from the energy the spring was compressed with.

1/2*(m+m_bul)*v^2 = 1/2*k*x^2

Solve for v and evaluate

1/2*(m+m_bul)*v^2 = 1/2*k*x^2
v^2 = k*x^2 / (m+m_bul)
v = sqrt (  k*x^2 / (m+m_bul) )
v = sqrt ( 3.972 N/m *  (0.18 m)^2 / (1 kg) )
v = sqrt ( 0.1287 m^2 / s^2 )
v = 3.587 m/s

Find the initial speed of the bullet
Because in an inelastic collision energy is not conserved - we must use the law of conservation of momentum.

m_bul * V_i + m * 0 = (m_bul + m) * v

solve for V_i

V_i = (m_bul + m) / m_bul * v
V_i = 1kg / 0.008 kg * 3.587 m/s = 448.375 m/s

Initial Speed of the Bullet = 448.375 m/s
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Added Wed, 10 Jun '15
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