Physics Question plz help!!!?

Question

The rocket of mass .05 kg is fired at 647 m/s north. Before the rocket is fired, the rocket launcher, which has a mass of 4.65 kg, is also moving north at 2 m/s. What is the velocity of the rocket launcher immediately after the rocket is launched assuming momentum is conserved?

Answer 1

Anakin_louix wrote:

"Well, here we must be careful :

The rocket launcher is moving to north WITH the rock, so, the total mass before the rocket was fired : 4.65 + 0.5 = 5.15 kg"

Anakin has misread the mass of the rocket to be 0.5 kg, when it is only 0.05 kg. Therefore:


The momentums before and AS the rocket was fired relate accordingly :

(total mass of rocket and rocket launcher as Anakin pointed out) X (initial velocity of rocket launcher) = [(mass of rocket)(new velocity of rocket)] + [(mass of rocket launcher)(new velocity of rocket launcher)]

4.7*2 = 0.05*647 + 4.65*Vlauncher (new)
9.4 = 32.35 + 4.65 kg * Vlauncher (new)
9.4 - 32.35 = 4.65*V
-22.95 = 4.65*V
Vlauncher (new) = V = -22.95/4.65 (m/s) = -4.94 m/s
(this was rounded up to the nearest hundreths m/s)

SO V = -4.94 m/s

This means that the rocket launcher is moving to the south at a new velocity of -4.94 m/s!

Which makes a whole lot more sense than -67.35 m/s.

Let us now examine why very briefly:

Common sense and knowledge of the principles/equations of conversation of momentum would tell us that 67.35 m/s is too large a number and if this number were the truth, well rocket launchers wouldn't work very well in real life.

Think of it this way:

The launcher weighs 4.65 kg which is 93 times greater than the weight of its rocket.

So if a force that allows the rocket to move at 647 m/s is exacted, shouldn't that force move something that weighs 93 times greater 93 times less? So a quick estimate of the new backward velocity or speed of the rocket launcher going south would be about 647 m/s / 93 = 6.96 m/s (rounded)

This is of course not totally correct, but it is what our intuition should tell us (this would be assuming the rocket launcher wasn't moving at all though to begin with).

The 2 m/s movement of the rocket launcher takes away from the effect of this force due to conversation of momentum, plus also, not all the energy from the rocket launch goes into the launcher, lots, most kinetic energy will go into the rocket itself..hopefully.

Therefore the answer of about -5 m/s makes a lot of sense if you think about it.

Any questions, comments, criticisms, compliments, doughnuts? Send them to me, via email.

Have a good day!

"Rocket Man"

Answer 2

Well, here we must be careful :

Yes I misread, the mass of the rock launcher is : 0.05 kg, so then :

Thanks to xza, his answer is the correct too

The rocket launcher is moving to north WITH the rock, so, the total mass before the rocket was fired : 4.65 + 0.05 = 4.7 kg

So, the momentum before the rocket was fired :

4.70*2 = 0.05*647 + 4.65*V

9.4 - 32.35 = 4.65*V

V = -4.90 m/s

minus in velocity means, that the rocket launcher is moving to the south.

Magnitude of the velocity = 4.90 m/s

Answer 3

State your givens:

m1 = .05 kg
Vo1 = 0.0 m/s
Vf1 = 647 m/s @north
m2 = 4.65 kg
Vo2 = 2 m/s @north
Vf2 = ?

Use this equation: po = pf
m1vo1 + m2vo2 = m1vf1 + m2vf2
m1vo1 = 0 so cancel that out

m2vo2 = m1vf1 + m2vf2

Isolate the variable you are looking for

(m2vo2 - m1vf1) / m2 = Vf2
(4.65kg * 2m/s - .05kg * 647m/s) / 4.65kg = Vf2
Vf2 = -4.956 m/s

The negative indicates it is moving opposite of the rocket, so south.

Vf2 = 5.0 m/s @ south

Hope that helped

Answer 4

i took physics, but that was 2 yrs ago in 6th gr. sorry i reelie wish i could help, but i hav short term memory [loss]. any class mates u can email or even ask over myspace or sumthin? if u do it over myspace, make sure the title is sumthin catchy like ima fight dis biitch n ppl will read it or if ur older im bankrupt or sumthin iono. make ppl wanna read it. hope i helped, but thats all i can think of bcuz i don remember how 2 do this kinda thing. sorry!