Question

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.

Answer 1

(a) Neglecting air resistance, calculate the shell’s velocity when it leaves
the mortar.

You can use conservation of energy

KE when it leaves the mortar = PE gained

1/2mv^2 = mgh

v^2 = 2gh

= 2(9.8)(110)

v = 46.43274706

= 46.43 m/s

(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).

Use the formula

v^2 = u^2 + 2as

v^2 = 0 + 2as

a = v^2/2s

= 46.43^2/2*0.45

= 2395.27211

= 2395.27 m/s^2


(c) What is the average force on the shell in the mortar?

F = ma

= 2.5 * 2395.27

= 5988.175 N

Weight of fireworks, W = 2.5 x 9.8

= 24.5 N

Ratio F : W = 5988.175/24.5

= 244.4153061 : 1

= 244 : 1