A 2.50 kg fireworks shell is fired straight up from a mortar and reaches a height of 110. m.
(a) Neglecting air resistance (a poor assumption, but we will make it for this example), calculate
the shell’s velocity when it leaves the mortar. (b) The mortar itself is a tube 0.450 m long.
Calculate the average acceleration of the shell in the tube as it goes from zero to the velocity found
in (a). (c) What is the average net force on the shell in the mortar? Food for thought: How does
this force compare to the weight of the shell?
a. vf^2=vi^2+2gh
vf=0, g=-9,8 solve for vi.
b. same equation as above, solve for a where g is.
c. Force*distance=1/2 m v^2
Thank you so much
To solve this problem, we need to use the principles of Newtonian mechanics. Let's break down the problem into three parts and solve them step-by-step.
(a) To calculate the shell's velocity when it leaves the mortar, we can use the principle of conservation of mechanical energy. At the top of its trajectory, the kinetic energy is zero, and the only energy the shell has is potential energy due to its height:
Potential Energy (PE) = mgh
where m is the mass of the shell (2.50 kg), g is the acceleration due to gravity (9.8 m/s^2), and h is the height reached by the shell (110 m).
PE = (2.50 kg) x (9.8 m/s^2) x (110 m) = 2695 J
At the start of its trajectory, the shell only has kinetic energy:
Kinetic Energy (KE) = (1/2)mv^2
where v is the velocity of the shell when it leaves the mortar (what we're trying to find).
Setting PE equal to KE:
2695 J = (1/2)(2.50 kg)(v^2)
Simplifying and solving for v:
v^2 = (2 * 2695 J) / (2.50 kg)
v^2 = 2156 J / (2.50 kg)
v^2 = 862.4 m^2/s^2
v = √(862.4 m^2/s^2)
v ≈ 29.4 m/s
Therefore, the shell's velocity when it leaves the mortar is approximately 29.4 m/s.
(b) To calculate the average acceleration of the shell in the tube, we can use the equation:
v = u + at
where v is the final velocity (29.4 m/s), u is the initial velocity (0 m/s), a is the average acceleration, and t is the time taken for the shell to reach the final velocity.
The length of the mortar is given as 0.450 m, so the distance traveled by the shell is 0.450 m. Using kinematic equations, we have:
v^2 = u^2 + 2as
Substituting the known values:
(29.4 m/s)^2 = (0 m/s)^2 + 2a(0.450 m)
862.4 m^2/s^2 = 0 + 2a(0.450 m)
Simplifying and solving for a:
a = (862.4 m^2/s^2) / (2 x 0.450 m)
a ≈ 959 m/s^2
Therefore, the average acceleration of the shell in the tube is approximately 959 m/s^2.
(c) To calculate the average net force on the shell in the mortar, we can use Newton's second law of motion:
F = ma
where F is the net force, m is the mass of the shell (2.50 kg), and a is the average acceleration (959 m/s^2).
F = (2.50 kg) x (959 m/s^2)
F ≈ 2397.5 N
Therefore, the average net force on the shell in the mortar is approximately 2397.5 N.
Food for thought: The weight of the shell can be calculated using the formula:
Weight = mg
where m is the mass of the shell (2.50 kg) and g is the acceleration due to gravity (9.8 m/s^2).
Weight = (2.50 kg) x (9.8 m/s^2)
Weight ≈ 24.5 N
Comparing the average net force on the shell in the mortar (2397.5 N) to its weight (24.5 N), we can see that the force exerted on the shell in the mortar is significantly greater than its weight.
To calculate the shell's velocity when it leaves the mortar, we can use the equations of motion. Let's assume the initial velocity of the shell when it is launched is zero. We can use the equation:
v² = u² + 2as
where
v = final velocity
u = initial velocity (zero in this case)
a = acceleration
s = displacement (height reached by the shell)
Given:
mass of the shell (m) = 2.50 kg
height reached (s) = 110 m
(a) Calculating the velocity when it leaves the mortar:
First, we need to find the value of acceleration. The only force acting on the shell is the gravitational force. So, we can use Newton's second law, F = ma, to find the acceleration. In this case, the only force acting on the shell is its weight, which can be calculated using the formula:
weight (W) = mass (m) * gravitational acceleration (g)
where
gravitational acceleration (g) = 9.8 m/s²
weight (W) = 2.50 kg * 9.8 m/s² = 24.5 N
Now, we can equate the weight to the net force acting on the shell (Fnet):
Fnet = W = mass (m) * acceleration (a)
24.5 N = 2.50 kg * a
a = 9.8 m/s²
Using the equation of motion:
v² = u² + 2as
v² = 0 + 2 * 9.8 m/s² * 110 m
v² = 2156 m²/s²
Taking the square root of both sides:
v = √(2156 m²/s²)
v ≈ 46.5 m/s
Therefore, the shell's velocity when it leaves the mortar is approximately 46.5 m/s.
(b) To find the average acceleration of the shell in the tube, we can use the equation:
acceleration (a) = (final velocity - initial velocity) / time
Given:
length of the tube (s) = 0.450 m
initial velocity (u) = 0 m/s
final velocity (v) = 46.5 m/s
Using the equation:
a = (v - u) / s
a = (46.5 m/s - 0 m/s) / 0.450 m
a ≈ 103.3 m/s²
Therefore, the average acceleration of the shell in the tube is approximately 103.3 m/s².
(c) To find the average net force on the shell in the mortar, we can use Newton's second law:
Fnet = mass (m) * acceleration (a)
Fnet = 2.50 kg * 103.3 m/s²
Fnet ≈ 258.2 N
Therefore, the average net force on the shell in the mortar is approximately 258.2 N.
As for the comparison between this force and the weight of the shell, we calculated the weight of the shell earlier to be 24.5 N. Therefore, the average net force on the shell in the mortar is significantly larger than its weight.