A 9.3 kg firework is launched straight up and at its maximum height 45 m it explodes into three parts. Part A (0.5 kg) moves straight down and lands 0.29 seconds after the explosion. Part B (1 kg) moves horizontally to the right and lands 10 meters from Part A. Part C moves to the left at some angle. How far from Part A does Part C land (no direction needed)?
I will be happy to critique your work, when shown. This is an exercise in applying Newton's second law.
1. Momentum for part A:
y = (Vo)t - (1/2)(9.8m/s^2)t^2
Substitute the values for y and t which are given. Solve for Vo, the initial downward velocity. Multiply the value of Vo by 0.5 kg to get the initial momentum down of A.
2. Momentum for part B:
y = (1/2)gt^2 = 45m for its downward motion.
substitute the value of g and solve for t.
Vx = 10m/t
Now you can calculate the momentum of B
3. Resultant of the two momentums:
Do a vector addition of the momentums of A and B. The resultant will be a vector downward to the right. Determine both, magnitude and direction.
4. Momentum of part C:
That is opposite (equilibrant) of of the vector sum of A and B. Determine it.
Divide the momentum of C by its mass to get the initial velocity of C. That will make part C a projectile at an upward angle. It will be tricky solving this part since it lands 45 meters below the point of "launching".
Best of luck.
To find the distance from Part A where Part C lands, we need to analyze the given information.
Let's calculate the initial vertical velocity of the firework before it explodes. We know that the firework reaches its maximum height and then falls down, so at the maximum height, the vertical velocity is zero.
Using the kinematic equation for vertical motion, which is:
vf = vi + at
where vf is the final velocity, vi is the initial velocity, a is the acceleration, and t is the time, we can solve for the initial vertical velocity.
At the maximum height, vf = 0, and let's assume that upward motion is positive and downward motion is negative.
So, 0 = vi + (-9.8 m/s^2) * t
Simplifying the equation, we have:
vi = 9.8 m/s * t
Given that the time at the maximum height is 45 m, we can substitute the values:
vi = 9.8 m/s * 0.29 s
vi = 2.842 m/s
Now, let's analyze the motion of Parts A, B, and C.
Part A has a mass of 0.5 kg, moves straight down, and lands 0.29 seconds after the explosion. We can use the equation for freely falling objects to calculate the distance it traveled:
d = vi * t + (1/2) * a * t^2
Since it moves straight down, and downward motion is negative, we can take a = -9.8 m/s^2:
dA = -(2.842 m/s) * (0.29 s) + (1/2) * (-9.8 m/s^2) * (0.29 s)^2
dA = -0.824 m
So, Part A lands 0.824 meters below the explosion point.
Part B has a mass of 1 kg, moves horizontally to the right, and lands 10 meters from Part A. As it moves horizontally, we know that there is no acceleration in the horizontal direction.
So, using the equation for constant velocity motion, we have:
dB = v * t
Since the horizontal motion is not affected by gravity or any other force, the horizontal velocity remains constant. Therefore, the horizontal velocity is equal to the initial horizontal velocity of the firework before it exploded.
Substituting the values, we have:
dB = 2.842 m/s * 0.29 s
dB = 0.825 m
Therefore, Part B lands 0.825 meters to the right of Part A.
Now, let's find the distance where Part C lands.
Since the firework exploded into three parts, the initial horizontal velocity is shared between Parts B and C.
The horizontal velocity of Part B is 0.825 m/s to the right, and the horizontal velocity of Part C can be considered as -vc since it moves to the left.
Using conservation of momentum in the horizontal direction:
(mB * vB) + (mC * vC) = 0
Since mB = 1 kg and vB = 0.825 m/s, we can solve for mC * vC:
mC * vC = -(mB * vB)
mC * (-vc) = -(1 kg * 0.825 m/s)
The masses cancel out, and we are left with:
vc = 0.825 m/s
Now, we can find the time it takes for Part C to land.
Using the equation for constant velocity motion in the horizontal direction, we have:
dc = vc * t
Since dc is the distance where Part C lands, we can plug in the values:
dc = 0.825 m/s * t
And we know that dc is the sum of the distances from Part A to Part C and from Part B to Part C:
dc = dB - dC
Substituting the values, we have:
0.825 m = dB - dc
Therefore, we can set up the equation:
0.825 m = 0.825 m/s * t - dc
Simplifying the equation, we get:
0.825 m + dc = 0.825 m/s * t
Now, we substitute dc = -0.824 m (as Part C lands below the explosion point) and solve for t:
0.825 m - 0.824 m = 0.825 m/s * t
0.001 m = 0.825 m/s * t
t = 0.001 m / 0.825 m/s
t = 0.0012 s
Now, we can find the distance where Part C lands by using the equation for constant velocity motion in the horizontal direction:
dc = vc * t
Substituting the values, we have:
dc = -0.825 m/s * 0.0012 s
dc ≈ -0.001 m
Therefore, Part C lands about 0.001 meters to the left of Part A.