A 0.0361-kg bullet is fired horizontally into a 2.74-kg wooden block attached to

Question

A 0.0361-kg bullet is fired horizontally into a 2.74-kg wooden block attached to one end of a massless, horizontal spring (k = 886 N/m). The other end of the spring is fixed in place, and the spring is unstrained initially. The block rests on a horizontal, frictionless surface. The bullet strikes the block perpendicularly and quickly comes to a halt within it. As a result of this completely inelastic collision, the spring is compressed along its axis and causes the block/bullet to oscillate with an amplitude of 0.279 m. What is the speed of the bullet?

Explanation / Answer

You know the masses of the bullet, m, and block, M. You also know the spring constant, k, and the oscillation distance of the block/bullet after collision, x. We want to solve for the bullet's velocity, vb. 2/3. Theory: This is an inelastic collision, so you know that kinetic energy is not conversed, but momentum is. We can define the momenta for the system on both sides of the collision. Likewise, after the collision, we know the physics of the block/bullet, more specifically, the energy of its motion and that it must become the potential energy of the spring. 4. For the conversation of momentum, we have: m*vb = (m+M)*vBB, where vBB is the velocity of the block/bullet. For the energy of the block/bullet system after the collision, we have (1/2)*(m+M)*vBB^2 = (1/2)*k*x^2. 5. Now you can solve for the speed of the bullet from the beginning by solving for vBB from the energy equation and substituting it into the momentum equation. Give it a try. I get vb = (x/m)*((m+M)*k)^(1/2). 6. Now plug in numbers. I don't have a calculator, so have a go at it.

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