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3 years ago

6

A 4.0 kg model rocket is launched, shooting 50.0 g of burned fuel from its exhaust at

1 answer:

Anettt [7]3 years ago

3 0

The velocity of the rocket is 7.8 m/s

Explanation:

We can solve the problem by using the law of conservation of momentum: in fact, in absence of external forces, the total momentum of the rocket+fuel system must be conserved.

Before the launch, the total momentum of the system is zero, since the rocket and the fuel are at rest:

$p=0$

After the launch, the total momentum is

$p=MV+mv$

where

M = 4.0 kg is the mass of the rocket

V is the velocity of the rocket

m = 50.0 g = 0.050 kg is the mass of the fuel ejected

v = -625 m/s is the velocity of the fuel (taking "backward" as negative direction)

Since the total momentum is conserved, we have

$0=MV+mv$

So we can solve the equation to find V, the velocity of the rocket:

$V=-\frac{mv}{M}=-\frac{(0.050)(625)}{4.0}=+7.8 m/s$

And the positive sign means the rocket moves forward.

Learn more about momentum:

brainly.com/question/7973509

brainly.com/question/6573742

brainly.com/question/2370982

brainly.com/question/9484203

#LearnwithBrainly

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3 years ago

Gennadij [26K]

Answer:

3.675 m

Explanation:

$a_{x} =0$ $v_{xo}=100$ $a_{y} =-g$ $v_{yo}=0$

X-direction | Y-direction

$R=x_{o}+ v_{xo} t$ | $y=y_{o}+v_{yo}t+\frac{1}{2}a_{y}t^2$

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