Q4.7-47PE - College Physics (Urone)

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.

Step-by-Step Solution

Answer

(a) The initial velocity of the player is 46.43 m/s.

(b) The acceleration is 2395.6 m/s².

(c) The force exerted on the shell in the mortar is 5989 N, and the ratio of force to weight is 244.5.

Step 1: Given data

• Height = 100 m.
• Mass of the shell = 2.50 kg.

Step 2: (a) Determine the initial velocity of the shell.

Apply the equation of motion as:

 

${\mathit{v}}^{\mathbf{2}}\mathbf{-}{\mathit{u}}^{\mathbf{2}}\mathbf{=}\mathbf{2}\mathit{g}\mathit{h}$

 

Here, v is the final speed, u is the initial speed, g is the acceleration due to gravity, and h is the height attained.

 

Substitute 0 for v, 110 m for h, and (-9.8) m/s² for g in the above expression, and we get,

 

$\begin{array}{c}0-u^2=2\times \left(-9.8\right){\text{\hspace{0.33em}m/s}}^2\times 110\text{\hspace{0.33em}m}\\ \text{u}=\sqrt{2156{\text{\hspace{0.33em}m}}^2{\text{/s}}^2}\\ u=46.43\text{\hspace{0.33em}m/s}\end{array}$

 

Hence, the initial velocity of the shell is 46.43 m/s.

Step 3: (b) Determine the acceleration during the straightening of the legs

Apply the equation of motion as:

 

${\mathit{v}}^{\mathbf{2}}\mathbf{-}{\mathit{u}}^{\mathbf{2}}\mathbf{=}\mathbf{2}\mathit{a}\mathit{s}$

 

Here, a is the acceleration, and s is the distance covered.

 

Substitute 0 for u, 46.43 m/s for v, and 0.45 m for s in the above expression, and we get,

 

$\begin{array}{c}{46.43}^2{\text{\hspace{0.33em}m}}^{\text{2}}{\text{/s}}^{\text{2}}-0=2\times a\times 0.45\text{\hspace{0.33em}m}\\ 2155.7{\text{\hspace{0.33em}m}}^{\text{2}}{\text{/s}}^{\text{2}}-0=a\times 0.9\text{\hspace{0.33em}m}\\ a=\frac{2155.7{\text{\hspace{0.33em}m}}^{\text{2}}{\text{/s}}^{\text{2}}}{0.9\text{\hspace{0.33em}m}}\\ a=2395.6{\text{\hspace{0.33em}m/s}}^2\end{array}$

 

Hence, the acceleration is 2395.6 m/s².

Step 4: (c) Determine the force exerted on the shell in the mortar

Apply Newton’s Second Law of motion as:

 

$F=ma$

 

Here, m is the mass of the player, and F  is the net force exerted.

 

Substitute 2.5 kg for m and 2395.6 m/s² for a in the above expression, and we get,

 

$\begin{array}{c}F=2.5\text{\hspace{0.33em}kg}\times {\text{2395.6\hspace{0.33em}m/s}}^2\\ F=5989\text{\hspace{0.33em}kg}\cdot {\text{m/s}}^2\\ F=5989\text{\hspace{0.33em}N}\end{array}$

 

Determine the weight of the shell as:

 

$w = mg$

 

Here, w is the weight of the shell.

 

Substitute 2.5 kg for m and 9.8 m/s² for g in the above expression, and we get,

 

$\begin{array}{c}\text{w}=\text{2.5\hspace{0.33em}kg}\times {\text{9.8\hspace{0.33em}m/s}}^2\\ =24.5\text{\hspace{0.33em}N}\end{array}$

 

Determine the ratio of force and weight as:

 

$\begin{array}{c}\frac{F}{w}=\frac{5989\text{\hspace{0.33em}N}}{24.5\text{\hspace{0.33em}N}}\\ =244.5\end{array}$

 

Hence, the ratio of force and weight is 244.5.