Question

A large rocket has a mass of 2.00×10⁶ kg at takeoff, and its engines produce a thrust of 3.50×10⁷ N.

(a) Find its initial acceleration if it takes off vertically.

(b) How long does it take to reach a velocity of 120 km/h straight up, assuming constant mass and thrust?

(c) In reality, the mass of a rocket decreases significantly as its fuel is consumed. Describe qualitatively how this affects the acceleration and time for this motion.

Short answer

(a)The initial acceleration of the rocket is 7.7 m/s².

(b) The time taken by the rocket is 4.329 s.

(c) When the mass of the rocket decreases, the acceleration of the rocket increases for the same value of thrust. With the increased acceleration, it takes less time than 4.329 s.

Step-by-step solution

Given data

• The mass of the rocket = 2.00×10⁶ kg.

• The thrust produced by engines = 3.50×10⁷ N.

(a) Determine the initial acceleration of the rocket.

Apply Newton’s Second Law of motion as:

$\begin{array}{c}{F}_{net}=ma\\ T-mg=ma\end{array}$

Here, m is the mass of the rocket, a is the acceleration, T is the engine thrust, g is the acceleration due to gravity, and F_net is the net force exerted.

Substitute 2x10⁶ kg for m, 3.5x10⁷ N forT, and 9.8 m/s² for g in the above expression, and we get,

$\begin{array}{c}3.5\times {10}^7\text{\hspace{0.33em}N}-2\times {10}^6\text{\hspace{0.33em}kg}\times {\text{9.8\hspace{0.33em}m/s}}^2=2\times {10}^6\text{\hspace{0.33em}kg}\times a\\ \left(3.5\times {10}^7-1.96\times {10}^7\right)\text{\hspace{0.33em}kg}\cdot {\text{m/s}}^{\text{2}}=2\times {10}^6\text{\hspace{0.33em}kg}\times a\\ a=\frac{1.54\times {10}^7\text{\hspace{0.33em}kg}\cdot {\text{m/s}}^{\text{2}}}{2\times {10}^6\text{\hspace{0.33em}kg}}\\ a=7.7{\text{\hspace{0.33em}m/s}}^2\end{array}$

Hence, the initial acceleration of the rocket is 7.7 m/s².

(b) Determine the time taken to reach the final velocity of 120 km/h.

Convert the speed from 120 km/h to m/s as:

$\begin{array}{c}120\text{\hspace{0.33em}km/h}=\text{120\hspace{0.33em}}\frac{\text{km}}{\text{h}}\times \frac{1000\text{\hspace{0.33em}m}}{1\text{\hspace{0.33em}km}}\times \frac{1\text{\hspace{0.33em}h}}{3600\text{\hspace{0.33em}s}}\\ =33.33\text{\hspace{0.33em}m/s}\end{array}$

Apply the equation of motion as:

$\mathit{v}\mathbf{=}\mathit{u}\mathbf{+}\mathit{a}\mathit{t}$

Here, v is the final speed, u is the initial speed, and t is the time taken.

Substitute 0 m/s for u, 7.7 m/s²for a, and 33.33 m/s for vin the above expression, and we get,

$\begin{array}{c}33.33\text{\hspace{0.33em}m/s}=0+7.7{\text{\hspace{0.33em}m/s}}^2\times t\\ t=\frac{33.33\text{\hspace{0.33em}m/s}}{7.7{\text{\hspace{0.33em}m/s}}^2}\\ t=4.329\text{\hspace{0.33em}s}\end{array}$

Hence, the time taken by the rocket is 4.329 s.

(c) Explanation of the effect of decreasing mass on acceleration and time.

When the mass of the rocket decreases, the acceleration of the rocket increases for the same value of thrust. With the increased acceleration, it takes less time than 4.329 s.