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OpenStudy (anonymous):

A Slingshot with a force constant of890N/m is used to propel a primitive 10 005kg starship into deep space by releasing a 5kg block of ice into space. How much should the slingshot be pulled in order to increase the ships speed by 5m/s? (unit is Gravitational potential energy)

OpenStudy (rajat97):

i didn't understand the significance of the kg block can you please explain me??

OpenStudy (anonymous):

neither do I, sorry

OpenStudy (rajat97):

okay i'll try to get it!

OpenStudy (anonymous):

the answer is 17m if that helps

OpenStudy (rajat97):

oh! i got that this is momentum conservation

OpenStudy (rajat97):

just one moment:)

OpenStudy (rajat97):

i'm trying it...

OpenStudy (rajat97):

i think energy conservation may be involved in this

OpenStudy (rajat97):

lol i'm getting (94)^1/2

OpenStudy (rajat97):

yes

OpenStudy (anonymous):

?

OpenStudy (rajat97):

i'm trying it even now this is damn good!

OpenStudy (anonymous):

yes, i know, im having difficulty figuring out what to start with :/

OpenStudy (anonymous):

what are you trying first? which equations?

OpenStudy (rajat97):

look i think we need to use two equations one is the equation for momentum conservation and another is th work energy theorem i think so!

OpenStudy (anonymous):

can we assume that the initial velocity of the starship is stationary?

OpenStudy (rajat97):

no

OpenStudy (gebooors):

Hello both of you! You are right, it is conservation of momentum and something more.

OpenStudy (anonymous):

Hello

OpenStudy (rajat97):

if we use these two equations , we can get the initial and the final velocities seperately i mean i got it seperately

OpenStudy (rajat97):

hello

OpenStudy (gebooors):

Speed is increased 5 m/s, so velocity may be different than zero

OpenStudy (rajat97):

yeah!

OpenStudy (anonymous):

doesn't the spring constant come into play too? would the equation \[F=Kx\] come into play?

OpenStudy (gebooors):

F= kx is necessary, good point

OpenStudy (rajat97):

i used these equations and the masses were so large that it was very hard to solve it so i took the help of an online calculator and the spring may come into the story in the work energy theorem or ..

OpenStudy (anonymous):

if we find force then we can find x, but im unsure how to find force

OpenStudy (rajat97):

but if we use momentum conservation , we will not get any term that would include force and the spring force is an internal force (due to which we can apply momentum conservation)

OpenStudy (gebooors):

You can calculate how much momentum of space ship Increases, but energy needs another equation.

OpenStudy (rajat97):

i didn't get that:)

OpenStudy (gebooors):

Momentum p =mv, both spaceship and ice block have same momentum, At the beginning p = 0.

OpenStudy (rajat97):

oh i got it now!

OpenStudy (gebooors):

I think rajat is working with energy.

OpenStudy (rajat97):

yes i am i'm actually working with both momentum conservation and energy conservation because according to my solution , i got two variables in n equaiton so i need both the equations what do you think?? is it right??

OpenStudy (gebooors):

Looks good, go on. This is somehow similar as radioactive decay. Kinetic energies of spaceship and ice block give energy equation.

OpenStudy (rajat97):

i think i got something..

OpenStudy (rajat97):

no i got nothing!

OpenStudy (gebooors):

Sorry, I have to log out, but go on working. I'll fan both of you, so lets keep in touch. But don't medals to me, nikkii, rajat did the work! Have a nice day!

OpenStudy (rajat97):

you helped us thanks a lot for giving your time for the question

OpenStudy (rajat97):

and have a nice day!

OpenStudy (anonymous):

THank you for your time!

OpenStudy (anonymous):

\[m _{1}v _{1o} + m _{2}v _{2o} = m _{1}v _{1f} + m _{2}v _{2f}\] \[\frac{ 1 }{2 } m _{1}v _{1o}^{2} + \frac{ 1 }{2 } m _{2}v _{2o} ^{2}= \frac{ 1 }{2 } m _{1}v _{1f}^{2} + \frac{ 1 }{2 } m _{2}v _{2f}^{2}\]

OpenStudy (anonymous):

are these the 2 equations you are using?

OpenStudy (rajat97):

the first one is correct and the second one is also correct but we need to bring the work done by the spring somewhere in the picture and these equations are absolutely correct!

OpenStudy (rajat97):

i'll try this one and get you on that in just 15 minutes:)

OpenStudy (anonymous):

thank you

OpenStudy (rajat97):

thanks for fanning me!

OpenStudy (anonymous):

no probs!

OpenStudy (gebooors):

I just remember that potential energy of the slingshot is E = 0.5kx^2. It is shared to spaceship and ice block. Can you calculate speed of the ice block?

OpenStudy (anonymous):

initial or final?

OpenStudy (anonymous):

how does that equation relate to speed?

OpenStudy (gebooors):

First ice block is moving with the spaceship, if it is moving. You know change of momentum of spaceship. It get 5 m/s speed more. Ice block is lighter, so it's speed is higher, and it comes From conservation of momentum

OpenStudy (anonymous):

what equation do we use to find change in momentum? we don't know the initial velocities or the final one of the ice block. Since the ice block is being released, would we get two different momentum's that add up to the original?

OpenStudy (gebooors):

Change of momentum of spaceship is p1= mv = 10005 kg * 5 m/s Then speed of ice block after slingshot is v2 = p1/ 5 kg

OpenStudy (rajat97):

what i did is i kept in mind the fact that internal forces cannot change the momentum of the system but they can change the kinetic energy of the system so i think the equation that nikkiii made for energy conservation needed something additional and that is the potential energy of the spring.

OpenStudy (anonymous):

how did you get the equations v2= p1/5kg?

OpenStudy (anonymous):

and wouldnt that be p2? since p2=mv=10005kg * 5m/s. We don't have the initial velocity of the starship yet

OpenStudy (gebooors):

Conservation of energy: Potential energy of slingshot = kinetic energy of spaceship + kinetic energy of ice block Problem is that kinetic energy is not shared equally.

OpenStudy (rajat97):

i wrote the momentum conservation equation as follows: let v1 be the speed of the spaceship initially so pi=10005v1 finally, i assumed the speed of the ice block to be v2 so the final momentum was pf=10000(v1 + 5)+5v2 (velocity of the 10000kg spaceship increases by 5m/s) so do you think that this is correct??

OpenStudy (gebooors):

Before slingshot situation is similar as in a gun. When a bullet has high speed, gun has a smaller recoil speed. Whether spaceship is moving or not before slingshot effect, is not important when working with momentum

OpenStudy (rajat97):

okay!

OpenStudy (rajat97):

i gotta go it's too late for me bye join you tomorrow:)

OpenStudy (anonymous):

bye

OpenStudy (gebooors):

According what rajat wrote, change of momentum is pf - pi But, let's continue tomorrow, I try to think this, too Bye to both of you, you were working hard.

OpenStudy (anonymous):

Bye!

OpenStudy (gebooors):

Hello , nikkii and rajat! I thought slingshot applies same force to spaceship and ice block. Forces are equal according the law of for action and reaction (Newton III law) Then W= Fx (work done by slingshot) We get W = 0,5kx^2 Then work done by slingshot is change of kinetic energy... I got 0.5kx^2 = 0.5mv^2 When you solve x, it is 16.76 m, could that be so simple? Maybe we thought this too difficulty

OpenStudy (anonymous):

Hey! and we were probably over analyzing it. Thank you Geboors!

OpenStudy (rajat97):

that really makes sense! you rock @Gebooors

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