Question

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.

Answer 1

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, $u=300\ m.s^{-1}$

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

$m.u=(M+m).v$

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

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

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

Answer 2

Answer: 19.5

Explanation: