Any Mechanical Engineers Among Us?

[dann2707]

This is a tough one!

A gun of mass 0.650 kg fires a bullet horizontally at a suspended target, in which the bullet becomes embedded. The gun recoils with a speed of 4.43 m s-1; the target has a mass of 2 kg and a velocity of 1.434 m s-1 just after impact.

Draw diagrams of the system before and just after the gun is fired, and before andjust after the bullet strikes the target, clearly stating

your sign convention. Calculate the mass of the bullet and its velocity.

Assume that the bullet does not lose any speed between exiting the gun barrel and hitting the target.

If someone could go through it step by step I sure would appreciate it

Many thanks, dan

[pogo]

i can't read ignore me.

Edited December 4, 2010 by pogo

[Leistonbmx]

Wow. Do you do this in classes etc? Got no notes?

Got some sort of idea where to start. But its a stupid time so i'm probably thinking wrong haha.

I'll be back later.

[dann2707]

Yep but it's a big step up from what we've been doing!

I know that momentum = mass x velocity

and in this instance the momentum of the recoil = momentum of the bullet

Just can't think where to go from there!

[RobinJI]

It's too late for me to actually go through the sums right now, but the key will the in the fact that the bullet gets embedded in the target.

So you know the bullets momentum is equal to the guns recoils momentum, and the targets momentum will also be equal to the bullets momentum.

So you know the targets momentum, it's speed, and how much it weighs, but you don't know the additional mass added to it by the bullet:

Guns mass x recoil speed = (targets mass + bullets mass) x targets velocity.

Bosh the numbers in and you'll find you've only got one variable left blank, rearrange the equation to find that and you know the bullets mass.

Once you know the bullets mass, you already know is momentum, so it's velocity should be easy enough.

Edited because I wrote momentum where I meant to say velocity.

PS, For the record in my somewhat tired state, I make the bullet 8.02 grams, travelling at 359.0399m/s, but they'll be more interested in how you got there than the answer.

Edited December 5, 2010 by RobinJI

[dann2707]

Brilliant stuff. Thanks for that. Just worked it out and it seems right.

[Tomm]

Once you know the mass of the bullet, you can also work out the velocity by using the kinetic energy equation (KE1=KE2) etc. Just if you wanted to do it differently to everyone else

EDIT: This doesn't seem to work, I get a bullet velocity of 22.7m/s. I'm not sure why. Anyone got any ideas?

KE(target) = 0.5 mv^2 = 0.5 * (mass of target + bullet) * v^2 = 0.5 (2+0.008) * 1.434^2 = 2.06458

KE (target) = KE (bullet before hitting target)

2.06458 = KE (bullet before hitting target) = 0.5 mv^2 = 0.5 * 0.008 * v^2

v^2 = 2.06458 / (0.5 * 0.008)

Which I get to be 22.7 m/s as I said.

The only reason I can think of is perhaps some of the energy got converted to heat or something? Either that or your lecturer invented a hypothetical scenario which only rings true for conservation of momentum but doesn't work for conservation of energy?

[dann2707]

Once you know the mass of the bullet, you can also work out the velocity by using the kinetic energy equation (KE1=KE2) etc. Just if you wanted to do it differently to everyone else

EDIT: This doesn't seem to work, I get a bullet velocity of 22.7m/s. I'm not sure why. Anyone got any ideas?

KE(target) = 0.5 mv^2 = 0.5 * (mass of target + bullet) * v^2 = 0.5 (2+0.008) * 1.434^2 = 2.06458

KE (target) = KE (bullet before hitting target)

2.06458 = KE (bullet before hitting target) = 0.5 mv^2 = 0.5 * 0.008 * v^2

v^2 = 2.06458 / (0.5 * 0.008)

Which I get to be 22.7 m/s as I said.

The only reason I can think of is perhaps some of the energy got converted to heat or something? Either that or your lecturer invented a hypothetical scenario which only rings true for conservation of momentum but doesn't work for conservation of energy?

To work out the bullets velocity after this I used the

m1v1 = m2v2 law

so 0.650kg x 4.43ms = 0.00802kg x v2

Then by re-arranging you get

0.650 x 4.43/0.00802kg = V2 (bullets velocity)

This works out at around 389ms (that figure was off top of my head)

[PaRtZ]

Like it. Lots of good logic and assumptions. Its been 8 months since ive done some proper work o my degree but that does ring a bell

[Tomm]

To work out the bullets velocity after this I used the

m1v1 = m2v2 law

Yeah I did it that way first (and agree with Robin's aswer although I think he has a rounding error ). But I always try and work things out a second way, if I can. It spices things up

[RobinJI]

Yeah I did it that way first (and agree with Robin's aswer although I think he has a rounding error ). But I always try and work things out a second way, if I can. It spices things up

Yeah, when I worked out the velocity I just used 0.00802kg, rather than the full un-rounded number, but I don't think you'd be marked down for that anyway, as long as it was shown in your working.

(PS, using the full un-rounded number I get it as 359.0611 (To 4dp), probably worth mentioning I'm using my phone not a propper calculator too )

(PPS, I can't get the kinetic energy way to work either, just doesn't like it.)

Edited December 7, 2010 by RobinJI

[Shaun H]

This is really nerdy but I absolutely love this type of question! (bullet and hanging target) haha If I'd got here sooner I'd have been all over it but it's already past my bedtime...

The one's where you have to calculate the angle of swing are the meatiest

[AdamR28]

Once you know the mass of the bullet, you can also work out the velocity by using the kinetic energy equation (KE1=KE2) etc. Just if you wanted to do it differently to everyone else

Yeah I would have done it that way too!

I'm confident you can always assume no losses unless the question specifies.

The one's where you have to calculate the angle of swing are the meatiest

Just adds one pretty simple step? But yeah, it's cool how you have to break the question down into loads of little parts.

[RobinJI]

Not sure why, but the kinetic energy equation defiantly doesn't work, always comes out at 20-something m/s.

Edited December 7, 2010 by RobinJI

[ManxTrialSpaz]

Not sure why, but the kinetic energy equation defiantly doesn't work, always comes out at 20-something m/s.

I think it's just because it's not an elastic collision as the two particles come together.

[Tomm]

I think it's just because it's not an elastic collision as the two particles come together.

Yeah I think this is right. It says the "bullet becomes embedded" which suggests some sort of deformation, I guess.

[AdamR28]

Hmm, I would have thought that was a 'you can use energy to work the answer out, since the bullet doesn't bounce off' type clue.

[ManxTrialSpaz]

Hmm, I would have thought that was a 'you can use energy to work the answer out, since the bullet doesn't bounce off' type clue.

I definitely see where you're coming from, but kinetic energy is only conserved in perfectly elastic collisions - collisions where particles rebound from each other. So if the question was altered to "the bullet rebounds after striking the target with a velocity of x", you'd find that there'd be conservation of both momentum and kinetic energy. Obviously in this scenario, the target velocity after impact would be different to the one stated.

Edited December 8, 2010 by ManxTrialSpaz