When a 60 g bullet is fired horizontally at a velocity of 600 m/s into a 3 kg block of soft wood that is initially at rest, a fascinating interaction occurs. The bullet penetrates the block, transferring some of its momentum to the block upon impact.
To understand this, we can apply the principle of conservation of momentum. Before the collision, only the bullet is moving, while the block is stationary. The total momentum of the system before the bullet hits the block can be calculated using the formula:
Momentum = mass × velocity
For the bullet, the momentum is:
p_bullet = 0.06 kg × 600 m/s = 36 kg·m/s
Since the block is at rest, its initial momentum is 0. Therefore, the total momentum before the bullet strikes the block is:
p_total_initial = p_bullet + p_block = 36 kg·m/s + 0 = 36 kg·m/s
After the bullet passes through the block, it retains some of its velocity. Let’s denote the final velocity of the bullet as vb2. The block now has some velocity as well, which we will denote as v_block. The momentum after the bullet emerges can be expressed as:
p_total_final = p_bullet_final + p_block_final
This can be rewritten as:
p_total_final = (0.06 kg × vb2) + (3 kg × v_block)
Using the conservation of momentum, we set the total initial momentum equal to the total final momentum:
36 kg·m/s = (0.06 kg × vb2) + (3 kg × v_block)
To solve for the final velocities, we need additional information about vb2 or the energy loss during the collision. However, what is clear is that the bullet will emerge with some reduced velocity, and the block will start moving with its own velocity as a result of the bullet’s impact.
This scenario highlights not only the conservation of momentum but also the energy transfer that occurs during the collision. The exact outcomes depend on the specifics of the interaction such as the materials involved, friction, and whether the bullet deforms or loses energy upon impact.