# Impulse - distance question [closed]

What impulse should be applied to an object of mass m, having a known coefficient of friction u to get to a distance d ?

Thanks!

Update 14.02.2011:

I still wasn't able to find an answer to this. Could you please share the solution? I need an "Idiot's solution" for this. I'm not experienced enough to find a formula by myself starting from your clues. I've searched the forums but they take into account other parameters. Thank you !

Update 13.02.2011:

Here is a representation of what I need: It's been a long time since I studied the principles of physics, sorry if I did not make myself clear in the first post.

I've tried formulas like:

1)m * a = u * m * g => a = u * g

2)a = v / t

3)d = v * t

So from these three equations, distance = u * g * t^2

The impulse p = m*v , so I get p= m * u * g * t

However, the time is not known and it is not important how long it takes for the object to get to target.

Does this formula (p= m * u * g * t) seem right? The mass, u, g are known, but what about time?

## closed as off-topic by tpg2114♦, Brandon Enright, Qmechanic♦Dec 3 '13 at 0:56

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• Can you find the force of friction? Are there other forces in the problem? Can you find the acceleration of the object while it is moving. Can you deduce the necessary initial velocity to arrive at the desired distance? What does that imply about the required impulse? – dmckee Feb 12 '11 at 23:42
• If you want our help, demonstrate what you've tried and exactly why it isn't working for you. – David Z Feb 13 '11 at 4:13
• A general comment I make to students is that "trying formulas" isn't the way to approach a physics problem. At least not one that take more than one step. Instead visualize the situation, and think what physics applies. The box will have to be sliding. There is friction, and that's a force so it causes an acceleration (constant or not?). Can you find the initial velocity necessary to get all the way? Can you relate velocity at the start to impulse, noting that impulse is roughly "how hard you hit it"? – dmckee Feb 15 '11 at 3:55

## 3 Answers

In terms of the equations that you have presented equn 1) is correct. The problem is with 2) and especially 3). The reason is that this object does not have a constant velocity during motion. So you will need to either remember or calculate the formulae for an object with an initial velocity and moving under acceleration. It is true that these formulae will also contain time t terms, but there are ways to eliminate the t term, which might occur once you see the corrected formula for motion under acceleration.

So to begin with the velocity of the object at time t is the initial velocity plus a term from the (constant) acceleration applied for time t.

Also I am not sure how much Calculus you remember. Equation 2) would be more accurate as $a=dv/dt$. With some basic calculus manipulations you can derive the equations. If you dont quite remember, then take a guess and try to confirm with some calculus you do remember or directly.

• ok, so instead of 3) I'll have: d = v0 * t + a * t^2 / 2. I suppose v0 = 0 so d = a * t^2/2 – shining-light Feb 13 '11 at 11:04
• I still don't understand how to get rid of time from these formulas. – shining-light Feb 13 '11 at 11:30
• Good progress! v0 is not zero as it is the initial velocity v following the Impulse. Also watch out for + - signs (for a) in applications here. As the object is slowing down (decelerating). I have found two different ways to eliminate t. The most intuitive is to add another formula for v(t) following my second para: again remembering that the object is slowing down under friction. So that formula gives a value for t when it stops (at D). – Roy Simpson Feb 13 '11 at 12:20
• @tanderson : Note that the Physics of this problem resembles (but is a little simpler than) traditional sport problems where a bat hits a ball. Initially the ball is stationary, then for a short time and distance the bat touches the ball. When the ball and bat separate the ball now has a non-zero velocity. Other forces (including friction forces) now take over. Eventually the ball stops some distance D after some time T. – Roy Simpson Feb 13 '11 at 14:04
• The velocity formula is: $v(t) = v_0 + at$ where v0 is the velocity after the Impulse. So we use this to calculate the time T via the fact that the box will have stopped at X. So v(T)=0 giving $T=-v_0/a$. Now put this T into the your new equation for d to get $d=-v_0^2/a + 1/2av_0^2/a^2=-1/2v_0^2/a$. But you already know that 1) $a=-g\mu$ (minus for directions). So $d=v_0^2/2g\mu$. Obtain v0 from this in terms of the variables. Now we still need to deal with the Impulse itself which determines v0 as $I=mv_0$ as you have also. This gives the full expression for I. – Roy Simpson Feb 14 '11 at 22:24

We assume the friction force is constant. Then the question is equivalent to, "What is the momentum of an object with mass $m$ as it strikes the ground if you drop it from height $d$, on a planet where gravity is $g/\mu$?

The energy of the object scales linearly with $m$, $d$, and gravity. The energy times twice the mass is the square of the momentum, so the momentum should scale as the square root of the distance dropped and square root of gravity, and should vary linearly in $m$.

What property of an object's motion is related to impulse? That is, if you know the impulse, and you know that the object started from rest, what can you conclude about its state of motion after the impulse?

The coefficient of friction has one primary purpose in life: to allow you to calculate the force of friction. What is the relationship between those two? Once you know the force, and the information that you know by answering the questions in the previous paragraph, try to chart a path from there to the answer you're looking for.

I can't speak for others here, but personally I'd be much more inclined to try to give you some more specific help if you show me something about the efforts you've already made.