Question

help please. Rocket problem; impulse, change in velocity

at t=0, a 2250kg in outer space fires an engine that exerts an increasing force on it in the +x-direction. This force obeys the equation
Fx=At2
where 1.25s is time, and has a magnitude of 781.25N when t=1.25s

1) Find the SI constant A.

2) What impulse does the engine exert on the rocket during the 1.50s interval starting 2.00s after the engine was fired?

3) By how much does the rocket's velocity change during this interval?

Answer 1

1) Find the SI constant A.

Fx = A(t^2)

t = 1.25 s => Fx = A(1.25^2) => A = Fx / (1.25^2) = 781.25N / (1.25^2) s^2 = 500N/s^2

A = 500 N/s^2


2) What impulse does the engine exert on the rocket during the 1.50s interval starting 2.00s after the engine was fired?

I = Integral of Fx * dt from t = 2.0 s to t = 3.5 s

I = Integral of A(t^2)dt from t = 2.0s to t = 3.5s

I = 500(t^3) / 3 from t = 2.0 s to t = 3.5s

I = [500/3]* [(3.5s)^3 - (2.0s)^3 ] = 5812.5 N*s



3) By how much does the rocket's velocity change during this interval?


I =  momentum change ; constant mass => m* velocity change =>

velocity change = I / m = 5812.5 N*s / 2250kg = 2.58 m/s

Answer 2

1) The value of the constant $A$ in SI units is $\boxed{500\,{{\text{N}} \mathord{\left/{\vphantom {{\text{N}} {{{\text{s}}^{\text{2}}}}}}\right.\kern-\nulldelimiterspace} {{{\text{s}}^{\text{2}}}}}}$.

2) The impulse imparted by engine on the rocket during $2.00\,{\text{s}}$ to $3.50\,{\text{s}}$ is $\boxed{5812.5\,{\text{N}} \cdot {\text{s}}}$.

3)  The change in velocity of the rocket during the interval $2.00\,{\text{s}}$ to $3.50\,{\text{s}}$ is  $\boxed{2.58\,{{\text{m}} \mathord{\left/{\vphantom{{\text{m}} {\text{s}}}} \right.\kern-\nulldelimiterspace} {\text{s}}}}$.

Explanation:

Part (1):

The force acting on the rocket is given as:

 ${F_x} = A{t^2}$

Since the magnitude of force experienced by the rocket is $781.25\,{\text{N}}$ at the instant $t = 1.25\,{\text{s}}$, the value of constant   $A$ is calculated as:

$\begin{aligned}781.25&= A{\left( {1.25} \right)^2}\\ A&= \frac{{781.25}}{{1.5625}}\\&= 500\,{{\text{N}} \mathord{\left/ {\vphantom {{\text{N}} {{{\text{s}}^{\text{2}}}}}} \right.\kern-\nulldelimiterspace} {{{\text{s}}^{\text{2}}}}}\\\end{aligned}$

Thus, the value of the constant   in SI units is $\boxed{500\,{{\text{N}} \mathord{\left/{\vphantom {{\text{N}}{{{\text{s}}^{\text{2}}}}}}\right.\kern-\nulldelimiterspace} {{{\text{s}}^{\text{2}}}}}}$.

Part (2):

The impulse imparted by the engine is the amount of force exerted in the small interval of time. The impulse imparted by the engine on the rocket during $2.00\,{\text{s}}$ to $3.50\,{\text{s}}$ is given by:

$I=\int\limits_{t = 2.0}^{3.5}{F \cdot dt}$

Substitute the expression of the force in the above expression of impulse.

$\begin{aligned}I&= \int\limits_{t = 2.0}^{3.5} {A{t^2}dt}\\&= A\left[ {\frac{{{t^3}}}{3}} \right]_{2.0}^{3.5}\\&= \frac{{500}}{3}\left[ {{{\left( {3.5}\right)}^3} - {{\left( {2.0} \right)}^3}} \right]\\&= 5812.5\,{\text{N}}\cdot {\text{s}}\\\end{aligned}$

Thus, the imparted by engine on the rocket during $2.00\,{\text{s}}$ to $3.50\,{\text{s}}$  is $\boxed{5812.5\,{\text{N}} \cdot {\text{s}}}$.

Part (3):

According to the impulse-momentum theorem, the impulse imparted by the engine is equal to the change in momentum of the rocket. It is expressed as:

$\begin{aligned}I&=\Deltap\hfill\\I&=m\times\Deltav \hfill\\\end{aligned}$

The mass of rocket is $2250\,{\text{kg}}$ and the impulse imparted by the engine is $5812.5\,{\text{N}} \cdot {\text{s}}$.

Substitute the values in the above expression.

$\begin{aligned}  5812.5 &= 2250\times\Delta v\\\Delta v&= \frac{{5812.5}}{{2250}}\,{{\text{m}}\mathord{\left/{\vphantom {{\text{m}} {\text{s}}}} \right.\kern-\nulldelimiterspace} {\text{s}}}\\&= 2.58\,{{\text{m}}\mathord{\left/{\vphantom{{\text{m}} {\text{s}}}} \right.\kern-\nulldelimiterspace} {\text{s}}}\\\end{aligned}$

Thus, the change in velocity of the rocket due to the impulse imparted by the engine is

$\boxed{2.58\,{{\text{m}} \mathord{\left/{\vphantom {{\text{m}} {\text{s}}}}\right.\kern-\nulldelimiterspace} {\text{s}}}}$

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Answer Details:

Grade: High School

Subject: Physics

Chapter: Impulse-momentum theorem

Keywords:  Impulse imparted, the engine exerts, 1.50s interval, starting at 2.00s, rocket’s velocity change, magnitude of force, 2250 kg, in outer space fires an engine, impulse-momentum theorem.