Question

You shoot a 0.0050-kg bullet into a 2.0-kg wooden block at rest on a horizontal surface. After hitting dead center on a hard knot that runs through the block horizontally, the bullet pushes out the knot. It takes the bullet 1.0 ms to travel through the block, and as it does so, it experiences an x component of acceleration of -4.8× 10^5 m/s2. After the bullet pushes the knot out, the knot and bullet together have an x component of velocity of +10 m/s. The knot carries 10% of the original inertia of the block.

What is the initial velocity of the bullet?

Using conservation of momentum, compute the final velocity of the block after the collision.

Calculate the initial kinetic energy of the block-knot-bullet system.

Calculate the final kinetic energy of the block-knot-bullet system.

Does the kinetic energy of the system change during the collision?

Which type of collision is it?

Answer 1

Inertia of the knot = 0.2 kg

We can perceive the system as bullet hitting block and knot and finally all three bodies move

We know that v = u-at

Since the bullet has a final velocity of v = 10m/s

u = 10+4.8*10^5*10^-3 = 490m/s

So Initial KE of system = 0.5*0.005*490^2 = 600.25 J

Now Using conservation of momentum

0.005*490 = 0.005*10+0.2*10+2*v

So v = 0.2m/s final velocity of block

So final KE = 0.5*(2*0.2^2 + 0.2*10^2 + 0.005*10^2) = 10.29 J

This is an inelastic collision