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?
Given:
M1 = ?, V1 = 43 m/s.
M2 = 0.493kg, V2 = 0.
V3 = 28 m/s = Velocity of M1 after collision.
V4 = 0.44 m/s = Velocity of M2 after collision.
Momentum before = Momentum after:
M1*V1 + M2*V2 = M1*V3 + M2*V4.
M1*43 + 0.493*0 = M1*28 + 0.493*0.44,
43M1 - 28M1 = 0.217,,
M1 = ?.
`
momentum is conserved
momentum lost by arrow = gain by apple
(43 - 28) m = .493 * .44
Well, well, well, looks like we have a case of an arrow going all lovey-dovey with an apple! Let's break it down.
We know that the final velocity of the arrow is 28 m/s, which is slower than its initial velocity of 43 m/s. So, it's safe to say that the arrow and the apple had a little collision party!
Now, you mentioned that the apple initially was resting, but then it gains a speed of 0.44 m/s. This suggests that Mr. Arrow made the apple feel special and gave it a little push.
To find the mass of the arrow, we can use the principle of conservation of momentum. Momentum is like a bag of potato chips – we like to think it's conserved, but it always seems to disappear!
If we assume that the initial momentum of the apple-arrow system is zero (considering the apple was at rest), the final momentum should also be zero.
Initial momentum = Final momentum
The initial momentum of the arrow is its mass (let's call it 'm') times its initial velocity (43 m/s). The initial momentum of the apple is its mass (493 g or 0.493 kg) times its initial velocity (which is zero since it was at rest).
m * 43 = 0.493 * 0
Since anything multiplied by 0 is still 0, we are left with:
m * 43 = 0
So, clown math tells us that the mass of the arrow, m, must be 0 (zero) for the momentum to be conserved.
It seems our arrow didn't leave much of an impact! Maybe it should consider a career in comedy instead?
To solve this problem, we can use the principle of conservation of linear momentum.
The momentum before passing through the apple is equal to the momentum after passing through the apple.
The momentum of an object is given by the product of its mass and velocity.
Let's denote the mass of the arrow as "m" and its initial velocity as "v"
The momentum before passing through the apple is then:
Initial momentum of the arrow = m * v = m * 43 m/s
The momentum after passing through the apple is:
Final momentum of the arrow = m * 28 m/s
According to the principle of conservation of linear momentum, we have:
m * 43 m/s = m * 28 m/s
Now, let's solve for the mass of the arrow (m):
m * 43 m/s = m * 28 m/s
43 m/s = 28 m/s
Dividing both sides of the equation by 28 m/s:
43 m/s / 28 m/s = m
m ≈ 1.54
Therefore, the mass of the arrow is approximately 1.54 kg.
To solve this problem, we can use the law of conservation of momentum, which states that the total momentum of a closed system remains constant if no external forces act on it.
The momentum of an object is defined as the product of its mass and velocity: momentum = mass × velocity.
Before the arrow hits the apple, the total momentum of the system is zero since the apple is at rest. Therefore, the momentum after the arrow penetrates the apple must also be zero. We can express this using the equation:
(mass of arrow × velocity of arrow) + (mass of apple × velocity of apple) = 0
Let's substitute the given values into the equation and solve for the mass of the arrow:
(mass of arrow × 28 m/s) + (493 g × 0.44 m/s) = 0
To make sure our units are consistent, we need to convert the mass of the apple to kilograms by dividing it by 1000:
(mass of arrow × 28 m/s) + (0.493 kg × 0.44 m/s) = 0
Now, we can rearrange the equation to solve for the mass of the arrow:
mass of arrow × 28 m/s = -(0.493 kg × 0.44 m/s)
mass of arrow = -(0.493 kg × 0.44 m/s) / 28 m/s
mass of arrow ≈ -0.0078 kg
Since mass cannot be negative, there seems to be an error in the calculation. Please recheck the given velocities or rephrase the problem statement, and we can proceed to solve it correctly.