A 4.40g bullet moving with an initial speed of vi = 405 m/s is fired into and passes through a 1.20 kg block.
The block, initially at rest on a frictionless horizontal surface, is connected to a spring with a spring constant of 880 N/m. If the block moves x = 4.37 cm to the right after impact, calculate the speed at which the bullet emerges from the block.
I don't see how this can be done without some crazy assumptions about energy lost. On the sliding, you have zero friction (no energy lost), but the block itself. Is one to assume no energy is lost in the block internally as a bullet goes through it? It can be worked with that assumption, but frankly, that is ridiculous.
I'm with Bob on this. The ridiculous answer goes something like:
1/2 (4.4e-3)405^2 - 1/2 880 (.0437^2) = 1/2 (4.4e-3) vf^2
But you've just torpedoed a whole generation of ballistic pendulum problems...
To calculate the speed at which the bullet emerges from the block, we can use the principle of conservation of momentum. According to this principle, the total momentum before the collision is equal to the total momentum after the collision.
Let's break down the problem into three phases:
Phase 1: Bullet moving toward the block
Phase 2: Bullet embedded in the block
Phase 3: Bullet and block system moving together
Let's calculate the momentum before and after each phase.
Phase 1:
The initial momentum of the bullet is given by the equation:
p1_initial = m1 * v1_initial
Where:
m1 = mass of the bullet = 4.40 g = 0.0044 kg (converted to kg)
v1_initial = initial speed of the bullet = 405 m/s
Phase 2:
Since the bullet enters and passes through the block, we can consider the bullet and the block as a single system. The momentum of the system is given by the equation:
p2_initial = (m1 + m2) * v2_initial
Where:
m2 = mass of the block = 1.20 kg
v2_initial = initial speed of the bullet + block system
Phase 3:
After the bullet is embedded in the block, both the bullet and the block move together.
The momentum of the bullet-block system after the collision is given by the equation:
p3_final = (m1 + m2) * v3_final
Where:
v3_final = final velocity of the bullet and block system
Now, let's calculate each phase and solve for v3_final.
Phase 1:
p1_initial = (0.0044 kg) * (405 m/s) = 1.782 kg·m/s
Phase 2:
p2_initial = (0.0044 kg + 1.20 kg) * v2_initial = (1.2044 kg) * v2_initial
Phase 3:
p3_final = (1.2044 kg) * v3_final
Since the total momentum before and after the collision is conserved, we have:
p1_initial = p3_final
1.782 kg·m/s = (1.2044 kg) * v3_final
Now, let's solve for v3_final:
v3_final = (1.782 kg·m/s) / (1.2044 kg)
v3_final ≈ 1.480 m/s
Therefore, the speed at which the bullet emerges from the block is approximately 1.480 m/s.