An arrow slows down from 43 m/s to 28 m/s as it passes through an apple. If the 493 g apple was originally at rest and sped up to 0.44, what must be the mass of the arrow?
see 12:21 entry
To solve this problem, we can use the principle of conservation of momentum. According to this principle, the total momentum before an event is equal to the total momentum after the event, assuming no external forces are acting on the system.
Let's break down the problem into two separate events: the arrow passing through the apple and the apple speeding up.
1. The momentum of the arrow before passing through the apple is given by the product of its mass (m_a) and initial velocity (v_a1):
Momentum before = m_a * v_a1
2. The momentum of the arrow after passing through the apple is given by the product of its mass (m_a) and final velocity (v_a2):
Momentum after = m_a * v_a2
3. The momentum of the apple after being hit by the arrow is given by the product of its mass (m_apple) and final velocity (v_apple2):
Momentum of apple = m_apple * v_apple2
Since the total momentum before and after the events should be the same, we can write the following equation:
Momentum before = Momentum after + Momentum of apple
Using the given values:
m_a * v_a1 = m_a * v_a2 + m_apple * v_apple2
Substituting the given values:
(m_a * 43) = (m_a * 28) + (0.493 * 0.44)
Simplifying the equation:
43m_a = 28m_a + 0.21752
Subtracting 28m_a from both sides:
15m_a = 0.21752
Dividing both sides by 15:
m_a = 0.0145 kg
Therefore, the mass of the arrow must be approximately 0.0145 kg.
Please note that this calculation assumes no energy loss in the collision between the arrow and the apple, which might not be the case in reality.