A bullet with mass 7.58 x 10^-3 kg and a horizontal velocity of 285 m/s strikes

Question

A bullet with mass 7.58 x 10^-3 kg and a horizontal velocity of 285 m/s strikes and embeds itself in a wooden block with mass 0.892kg. The block rests on a frictionless surface and is attatched to one end of a spring. The other end of the spring is attached to the wall. The impact compresses the spring a maxium distance of .0180m. After impact, the block moves in simple harmonic motion.

a) Find the velocity of the block just after the bullet embeds itself in it.

b) Find the period of the simple harmonic motion that results from this process.

c) If the bullet strikes the block at t = 0, what is the speed and direction of the block at t = 2.10 seconds?

Please show work.

Explanation / Answer

Conservation of momentum:

m1u1 + m2u2 = (m1 + m2)v

v = 7.58 * 10^-3 * 285 / (0.892 + 0.00758)

= 2.40 m/s

V = 2.40 m/s

Kinetic energy is converted to potential energy

thus, 0.5 * m * v^2 = 0.5 * k * x^2

k = 15992.53 N/m

Period = 2*pi*sqrt(m/k) = 0.0471 s

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