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 m/s, what is the mass of the arrow?
Answer is 14.5 g
To solve this problem, we can use the principle of conservation of momentum. The total momentum before the collision is equal to the total momentum after the collision.
The momentum of an object is given by the equation: momentum = mass × velocity.
Let's assume that the mass of the arrow is "m" grams.
Before the collision:
The momentum of the arrow is given by: momentum_arrow = mass_arrow × velocity_arrow = m × 43 m/s.
The momentum of the apple is given by: momentum_apple = mass_apple × velocity_apple = 493 g × 0 m/s (since it was originally at rest).
After the collision:
The momentum of the arrow is given by: momentum_arrow = mass_arrow × velocity_arrow = m × 28 m/s.
The momentum of the apple is given by: momentum_apple = mass_apple × velocity_apple = 493 g × 0.44 m/s.
According to the principle of conservation of momentum, the total momentum before the collision is equal to the total momentum after the collision:
(m × 43 m/s) + (493 g × 0 m/s) = (m × 28 m/s) + (493 g × 0.44 m/s)
Simplifying the equation:
43m = 28m + 217.12
Subtracting 28m from both sides:
15m = 217.12
Dividing both sides by 15:
m = 14.4747...
Rounded to one decimal place, the mass of the arrow is approximately 14.5 g.