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3 years ago

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A rocket takes off from Earth and reaches a speed of 105 m/s in 18.0 s. If the exhaust speed is 1,200 m/sand the mass of fuel bu

rned is 110 kg, what was the initial mass (in kg, including the initial fuel) of the rocket?

1 answer:

Amanda [17]3 years ago

7 0

Answer:

The initial mass of the rocket is 526.2 kg.

Explanation:

Given that,

Final speed = 105 m/s

Time = 18.0 s

Exhaust initial speed = 1200 m/s

Mass of burned fuel = 110 kg

We need to calculate the initial mass

The velocity change of rocket under gravity is defined as,

$v=u\ ln(\dfrac{m_{i}}{m})-gt$ ....(I)

We know that,

$m=m_{i}-m_{bf}$

Put the value of m in equation (I)

$v=u\ ln(\dfrac{m_{i}}{m_{i}-m_{bf}})-gt$

$m_{i}=\dfrac{m_{bf}}{1-e^-{\dfrac{v+gt}{u}}}$

$m_{i}=\dfrac{110}{1- e^-{\frac{105+9.8\times18}{1200}}}$

$m_{i}=526.2 kg$

Hence, The initial mass of the rocket is 526.2 kg.

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Suppose a rocket ship accelerates upwards with acceleration equal in magnitude to twice the magnitude of g (we say that the rock

pashok25 [27]

Answer:

a) $s_a=98100\ m$ is the height where the rocket stops accelerating and its fuel is finished and starts decelerating while it still continues to move in the upward direction.

b) $v_a=1962\ m.s^{-1}$ is speed of the rocket going when it stops accelerating.

c) $H=294300\ m$

d) $t_T=544.95\ s$

e) Zero, since the average velocity is the net displacement per unit time and when the rocket strikes back the earth surface the net displacement is zero.

Explanation:

Given:

acceleration of rocket, $a=2g=2\times 9.81=19.62\ m.s^{-2}$

time for which the rocket accelerates, $t_a=100\ s$

<u>For the course of upward acceleration:</u>

using eq. of motion,

$s_a=ut+\frac{1}{2}at_a^2$

where:

$u=$ initial velocity of the rocket at the launch $=0$

$s_a=$ height the rocket travels just before its fuel finishes off

so,

$s_a=0+\frac{1}{2}\times 19.62\times 100^2$

a) $s_a=98100\ m$ is the height where the rocket stops accelerating and its fuel is finished and starts decelerating while it still continues to move in the upward direction.

<u>Now the velocity of the rocket just after the fuel is finished:</u>

$v_a=u+at_a$

$v_a=0+19.62\times 100$

b) $v_a=1962\ m.s^{-1}$ is speed of the rocket going when it stops accelerating.

After the fuel is finished the rocket starts to decelerates. So, we find the height of the rocket before it begins to fall back towards the earth.

Now the additional height the rocket ascends before it begins to fall back on the earth after the fuel is consumed completely, at this point its instantaneous velocity is zero:

using equation of motion,

$v^2=v_a^2-2gh$

where:

$g=$ acceleration due to gravity

$v=$ final velocity of the rocket at the top height

$0^2=1962^2-2\times 9.81\times h$

$h=196200\ m$

c) So the total height at which the rocket gets:

$H=h+s$

$H=196200+98100$

$H=294300\ m$

d)

Time taken by the rocket to reach the top height after the fuel is over:

$v=v_a+g.t$

$0=1962-9.81t$

$t=200\ s$

Now the time taken to fall from the total height:

$H=v.t'+\frac{1}{2}\times gt'^2$

$294300=0+0.5\times 9.81\times t'^2$

$t'=244.95\ s$

Hence the total time taken by the rocket to strike back on the earth:

$t_T=t_a+t+t'$

$t_T=100+200+244.95$

$t_T=544.95\ s$

e)

Zero, since the average velocity is the net displacement per unit time and when the rocket strikes back the earth surface the net displacement is zero.

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