Final answer:
By using the principle of conservation of momentum, we determine that the speed of the block and bullet when they first start moving together after the collision is approximately 2.97 m/s. Since this is not an option, the closest answer provided is (C) 2.94 m/s.
Step-by-step explanation:
To find the speed of the block and bullet when they first start moving together after the collision, 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, provided no external forces act on the system.
Let's calculate the momentum before the collision: The bullet's mass is 0.02 kg, and its velocity is 300 m/s. The momentum of the bullet (p_bullet) is mass times velocity:
p_bullet = 0.02 kg × 300 m/s = 6 kg·m/s
The block is initially at rest, so its momentum is zero. Therefore, the total momentum before the collision (p_total) is equal to the momentum of the bullet.
p_total = p_bullet + p_block_initial = 6 kg·m/s
After the collision, the bullet and block move together with a common velocity (v_common). Let's denote the combined mass of the block and bullet as m_combined:
m_combined = 2 kg + 0.02 kg = 2.02 kg
Using conservation of momentum, we have:
m_bullet × v_bullet + m_block_initial × v_block_initial = m_combined × v_common
Plug in the known values:
0.02 kg × 300 m/s + 0 kg × 0 m/s = 2.02 kg × v_common
From this, we can solve for v_common:
v_common = (0.02 kg × 300 m/s) / 2.02 kg = 6 kg·m/s / 2.02 kg ≈ 2.97 m/s
However, since we don't have 2.97 m/s as an option, the closest answer would be (C) 2.94 m/s.