A 0.05 kg arrow with the velocity of 180 m/s is used to shoot a 5 kg block of wood off the top of a log. What will be the velocity of the arrow and wood after contact?
To answer this question, we can use the principle of conservation of momentum. According to this principle, the total momentum before and after an interaction remains constant, assuming no external forces act on the system.
The momentum of an object is calculated by multiplying its mass by its velocity. Let's first find the momentum of the arrow before the interaction:
Momentum of arrow before = mass of arrow × velocity of arrow
= 0.05 kg × 180 m/s
= 9 kg·m/s
Now, since the arrow hits the block of wood, we can consider the arrow and the wood as a system. The total momentum of the system before the interaction is the sum of the arrow's momentum and the wood's momentum, which is initially at rest:
Total momentum before = momentum of arrow before + momentum of wood before
= 9 kg·m/s + 0 kg·m/s (wood is at rest)
= 9 kg·m/s
Since no external forces act on the system, the total momentum after the interaction should also be 9 kg·m/s.
Let's denote the final velocity of the arrow and wood as V. Now we can write the conservation of momentum equation:
Total momentum after = (mass of arrow + mass of wood) × final velocity
= (0.05 kg + 5 kg) × V
= 5.05 kg × V
Since the total momentum after the interaction is 9 kg·m/s, we can set up the equation:
5.05 kg × V = 9 kg·m/s
Now, we can solve for the final velocity, V:
V = 9 kg·m/s / 5.05 kg
V ≈ 1.782 m/s
Therefore, the velocity of the arrow and wood after contact is approximately 1.782 m/s.