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A 2.50 kg fireworks shell is fired straight up from a mortar and reaches a height of 110. m. (a)...

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 net force on the shell in the mortar? How does this force compare to the weight of the shell?

Acceleration:

Assume a particle on which a force is imposed and the velocity of the particle changes constantly. In this scenario, the rate of change of velocity for the particle would be termed as its acceleration.

Answer and Explanation:

Given data:

  • Mass of the shell, {eq}m = 2.50 \ kg {/eq}
  • Height, {eq}h = 110 \ m {/eq}
  • Length of the barrel, {eq}d = 0.450 \ m {/eq}
  • Initial speed of the shell in the barrel, {eq}u = 0 {/eq}

Part (a):

Let the lauched speed of the sheel be v.

From the conservation law o mechanical energy,

{eq}\begin{align*} \frac{1}{2}mv^{2} &= mgh\\ \Rightarrow \ v &= \sqrt{2gh}\\ v &= \sqrt{2 \times 9.80 \times 110}\\ v &= 46.43 \ m/s.\\ \end{align*} {/eq}

Part (b):

The average acceleration of the sheel can be given as,

{eq}\begin{align*} a &= \frac{v^{2}-u^{2}}{2d}\\ a &= \frac{46.43^{2}-0^{2}}{2 \times 0.450}\\ a &= 2395.27 \ m/s^{2}.\\ \end{align*} {/eq}

Part (c):

The average net force on sheel can be given as,

{eq}\begin{align*} F &= ma\\ F &=2.50 \times 2395.27 \\ F &= 5988.18 \ \rm N.\\ \end{align*} {/eq}

On comparing this force with the weight of the sheel,

{eq}\begin{align*} \frac{F}{W} &= \frac{5988.18 }{2.50 \times 9.80}\\ F &= 244.41 W.\\ \end{align*} {/eq}


Learn more about this topic:

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What is Acceleration? - Definition and Formula

from CLEP Natural Sciences: Study Guide & Test Prep

Chapter 4 / Lesson 2
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