Calculate the speed of a proton given its kinetic energy

[Hannah1]

Calculate the speed of a proton given its kinetic energy!

Homework Statement

Find the speed of a proton (mass = 1.67E-27 kg), if its kinetic energy is 145eV. Write your answer correct to three significant figures.

Homework Equations

W=.5*m*v^2
Proton Mass = 1.67E-27 kg

I tried using the same method used for the problem on this link:
http://tuhsphysics.ttsd.k12.or.us/Tutorial/NewIBPS/PS17/PS17.htm#p5

The Attempt at a Solution

145 eV = 2.32315587 × 10^-17 joules

W=.5 * m * v^2
2.32315587E-17 = .5 * 1.67E-27 * v^2
2.32315587E-17 / .5 / 1.67E-27 = v^2
v^2 = 27822225988 = 2.7822226E10
v = 166799.957998 = 1.66799957998E5 = 1.67E5

= 1.67E5 m/s

This was my answer on the homework, but I got it wrong.
What did I do wrong??

 

[Dick]

Homework Statement

Find the speed of a proton (mass = 1.67E-27 kg), if its kinetic energy is 145eV. Write your answer correct to three significant figures.

Homework Equations

W=.5*m*v^2
Proton Mass = 1.67E-27 kg

I tried using the same method used for the problem on this link:
http://tuhsphysics.ttsd.k12.or.us/Tutorial/NewIBPS/PS17/PS17.htm#p5

The Attempt at a Solution

145 eV = 2.32315587 × 10^-17 joules

W=.5 * m * v^2
2.32315587E-17 = .5 * 1.67E-27 * v^2
2.32315587E-17 / .5 / 1.67E-27 = v^2
v^2 = 27822225988 = 2.7822226E10
v = 166799.957998 = 1.66799957998E5 = 1.67E5

= 1.67E5 m/s

This was my answer on the homework, but I got it wrong.
What did I do wrong??

It looks ok to me.

 

[Hannah1]

There must be another answer though because other people got the question correct.

 

[Dick]

There must be another answer though because other people got the question correct.

Could you ask them what they answered? If it makes you feel any better I would have gotten it 'wrong' too. Oddball grading criteria can defeat the best of us.

 

[Curious3141]

Homework Statement

Find the speed of a proton (mass = 1.67E-27 kg), if its kinetic energy is 145eV. Write your answer correct to three significant figures.

Homework Equations

W=.5*m*v^2
Proton Mass = 1.67E-27 kg

I tried using the same method used for the problem on this link:
http://tuhsphysics.ttsd.k12.or.us/Tutorial/NewIBPS/PS17/PS17.htm#p5

The Attempt at a Solution

145 eV = 2.32315587 × 10^-17 joules

W=.5 * m * v^2
2.32315587E-17 = .5 * 1.67E-27 * v^2
2.32315587E-17 / .5 / 1.67E-27 = v^2
v^2 = 27822225988 = 2.7822226E10
v = 166799.957998 = 1.66799957998E5 = 1.67E5

= 1.67E5 m/s

This was my answer on the homework, but I got it wrong.
What did I do wrong??

Is your answer marked wrong, or the method?

It is possible your teacher may expect you to use Special Relativity to get the answer. But at the energy levels given, the answer will not be significantly different.

 

[Dick]

Is your answer marked wrong, or the method?

It is possible your teacher may expect you to use Special Relativity to get the answer. But at the energy levels given, the answer will not be significantly different.

Well the relativistic answer might not affect the answer significantly. But it might affect a three decimal point round-off. I'll emphasize MIGHT. I'm not even going to work it out. This doesn't sound like a course that does relativisitic corrections.

 

[Mute]

Well the relativistic answer might not affect the answer significantly. But it might affect a three decimal point round-off. I'll emphasize MIGHT. I'm not even going to work it out. This doesn't sound like a course that does relativisitic corrections.

I checked the relativistic formula earlier; it gets the same answer to the desired number of significant figures (so I didn't make a post).

 

[Curious3141]

I checked the relativistic formula earlier; it gets the same answer to the desired number of significant figures (so I didn't make a post).

I checked as well (before my post).

 

[Dick]

I checked as well (before my post).

You guys are thorough. Good job. Did you check rounding off all of the intermediate results to three decimal points? I tried that. Still didn't work. Or rather did work, in the sense I got the same answer.

 

[Curious3141]

You guys are thorough. Good job. Did you check rounding off all of the intermediate results to three decimal points? I tried that. Still didn't work.

I feel a bit shame-faced because I used c = 3*10⁸ m/s. I was rushing for time.

But not shame-faced enough that I'm going to go back and check again.

I can, however, help the OP to do the checking himself/herself. The SR formula (after algebraic manipulation to isolate v) is:

$$v = c\sqrt{1 - {(\frac{K}{mc^2} + 1)}^{-2}}$$

 

[Dick]

I feel a bit shame-faced because I used c = 3*10⁸ m/s. I was rushing for time.

But not shame-faced enough that I'm going to go back and check again.

I can, however, help the OP to do the checking himself/herself. The SR formula (after algebraic manipulation to isolate v) is:

$$v = c\sqrt{1 - {(\frac{K}{mc^2} + 1)}^{-2}}$$

We all know that none of these reasonable suggestions are really the answer, right? The homework grading program is broken. Uh, just try rounding down even though it would be wrong. Might work. Uh, try anything.

 

[Curious3141]

We all know that none of these reasonable suggestions are really the answer, right?

Well, as the problem is stated, what we (including the OP) got IS the correct answer. If the expected answer differs by anything other than a 'technicality' (round-off error, expectation of SR method, etc.), then the expected answer is wrong.

 

[Dick]

Well, as the problem is stated, what we (including the OP) got IS the correct answer. If the expected answer differs by anything other than a 'technicality' (round-off error, expectation of SR method, etc.), then the expected answer is wrong.

That's what I meant to say. Sure. It is wrong. Everybody else in the room is right. I just meant I was trying to get some other answer besides 1.67E5 m/s by rigging the system in any possible way. And I failed. Well, I can get 1.66E5 m/s by rounding down. I.e. just looking at the digits showing on the calculator and ignoring the rest.

 

[Curious3141]

That's what I meant to say. Sure. It is wrong. Everybody else in the room is right. I just meant I was trying to get some other answer besides 1.67E5 m/s by rigging the system in any possible way. And I failed.

Aw, heck, I plugged all the "exact" values (as google has them anyway) into the SR formula using google calculator and the value I got was:

$$v = 166669.170677 = 1.67 \times 10^{5} ms^{-1}$$

to the required number of sig figs.

 

[Dick]

Aw, heck, I plugged all the "exact" values (as google has them anyway) into the SR formula using google calculator and the value I got was:

$$v = 166669.170677 = 1.67 \times 10^{5} ms^{-1}$$

to the required number of sig figs.

Ok, so we have all failed to get anything different from what the OP got, right? Somebodies got some splaining to do. There's more of these 'right answers ruled wrong for no apparent reason' on the Hannah1 line.

 

[Mute]

I feel a bit shame-faced because I used c = 3*10⁸ m/s. I was rushing for time.

I just typed "(speed of light)*(<rest of expression with numbers>)" into google. =P

Aw, heck, I plugged all the "exact" values (as google has them anyway) into the SR formula using google calculator and the value I got was:

$$v = 166669.170677 = 1.67 \times 10^{5} ms^{-1}$$

to the required number of sig figs.

Yep, that's what I got earlier. =P I used the proton mass in GeV/c², too.

Hannah1, are you absolutely 140% sure you're using the right numbers for the Kinetic energy? It's not supposed to be 145 keV or anything? Or numerically different? Otherwise, yeah, seems like your homework system might have a bug. =/

 

[Hannah1]

Thank you so much everyone for helping me!

The teacher doesn't grade the method we use because we submit everything online. She gives us 3 chances to complete it so we can better our grade.

I tried again with the homework earlier today, but still got the answer wrong.

-------------------------------------------------------------

Here is the problem on my second attempt:

Find the speed of a proton (mass = 1.67E-27 kg), if its kinetic energy is 133eV. Write your answer correct to three significant figures.
(Note: The teacher changes the value of the numbers in the problem for each HW attempt...)

133 eV = 2.13089469 × 10^-17 joules

W = .5 * m * v²
2.13089469E-17 = .5 * 1.67E-27 * v²
2.13089469E-17 / .5 / 1.67E-27 = v² (On my second attempt, I tried dividing two numbers at a time instead of all three.)
4.2617894E-17 / 1.67E-27 = v² (On this step, I divided the last two numbers.)
v² = 25519697006
v = 159748.85604

= 1.60E5 m/s

I submitted my answer, and I still got it counted as incorrect.
Please help, I only have one more attempt!

 

[Hannah1]

The only things that I can think of are that W = .5 * m * v2 might be the wrong formula to use... or that m/s might be the wrong unit to use.

Other than that, I have no idea!

 

[gneill]

Make sure that the format you supply matches the format expected by the system (particularly for exponents). Double check the units that they want the answer provided in (watch out for things like "answer in units of c", "answer in km/hr).