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Bond [772]
3 years ago
9

A 10 g bullet is fired into, and embeds itself in, a 2 kg block attached to a spring with a force constant of 19.6 n/m and whose

mass is negligible. How far is the spring compressed if the bullet has a speed of 300 m/s just before it strikes the block and the block slides on a frictionless surface? Use conservation of momentum.
Physics
2 answers:
dmitriy555 [2]3 years ago
4 0

Answer:

x=0.478\ m is the compression in the spring

Explanation:

Given:

  • mass of the bullet, m=10\ g=0.01\ kg
  • mass of block, M=2\ kg
  • stiffness constant of the spring, k=19.6\ N.m^{-1}
  • initial velocity of the spring just before it hits the block, u=300\ m.s^{-1}

<u>Now since the bullet-mass gets embed into the block, we apply the conservation of momentum as:</u>

m.u=(M+m).v

0.01\times 300=(2+0.01)\times v

v=1.4925\ m.s^{-1}

Now this kinetic energy of the combined mass gets converted into potential energy of the spring.

\rm Kinetic\ energy=Spring\ potential\ energy

\frac{1}{2} (M+m).v^2=\frac{1}{2} k.x^2

\frac{1}{2} \times (2+0.01)\times 1.4925^2=\frac{1}{2} \times 19.6\times x^2

x=0.478\ m is the compression in the spring

Valentin [98]3 years ago
3 0

Answer: 19.5

Explanation:

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