Question

Integrated Concepts 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 force on the shell in the mortar? Express your answer in newtons and as a ratio to the weight of the shell.Neglecting air resistance, calculate the fireworks shell's velocity when leaving the mortar, reaching a height of 110 m.

a) 46.9 m/s
b) 52.3 m/s
c) 58.7 m/s
d) 64.2 m/s

Answer 1

(a) The shell's velocity when it leaves the mortar is approximately 46.5 m/s. (b) The average acceleration of the shell in the tube is approximately 103.3 m/s². (c) The average force on the shell in the mortar is approximately 258.3 Newtons.

Step-by-step explanation:

(a) To calculate the shell's velocity when it leaves the mortar, we can use the equation:

v^2 = u^2 + 2as

Where v is the final velocity (which is the velocity when the shell leaves the mortar), u is the initial velocity, a is the acceleration (which is equal to -g, the acceleration due to gravity), and s is the displacement (which is equal to the height reached by the shell).

Using the given values, we have:

v^2 = 0^2 + 2*(-9.8)*110

v^2 = 2156

v = √(2156)

v ≈ 46.5 m/s

So, the velocity when the shell leaves the mortar is approximately 46.5 m/s.

(b) To calculate the average acceleration of the shell in the tube, we can use the equation:

a = (v - u) / t

Where a is the average acceleration, v is the final velocity, u is the initial velocity (which is 0, as the shell starts from rest), and t is the time taken to reach the final velocity.

Using the values from part (a), we have:

a = (46.5 - 0) / (0.450)

a ≈ 103.3 m/s²

So, the average acceleration of the shell in the tube is approximately 103.3 m/s².

(c) The average force on the shell in the mortar can be calculated using Newton's second law of motion:

F = m * a

Where F is the force, m is the mass of the shell, and a is the acceleration.

Using the given mass of the shell (2.50 kg) and the calculated acceleration from part (b), we have:

F = 2.50 * 103.3

F ≈ 258.3 N

So, the average force on the shell in the mortar is approximately 258.3 Newtons.