Question

A 4.0 kg model rocket is launched, shooting 50.0 g of burned fuel from its exhaust at
an average velocity of 625 m/s. What is the velocity of the rocket after the fuel has
burned? •
Thiss is the answer I just need to show work>>>>>>>(7.8 m/s)

Answer 1

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.

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Answer 2

The velocity of the rocket after the fuel has burned is 7.81 m/s.          

From law of conservation of Momentum,

                         $MV=mv$

Where M is mass of rocket and V is velocity of rocket

And m is mass of burned fuel and v is speed of fuel.

Given that, $M=4Kg,m=50g=0.05Kg,v=625m/s$

Substitute values in above equation,

                  $4*V=0.05*625\\\\V=\frac{0.05*625}{4} =7.81m/s$

Hence, the velocity of the rocket after the fuel has burned is 7.81 m/s.              

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