Can You Solve This Motion Problem Using a Simple Trick?

[rudransh verma]

No, while falling through the air there is one force, gravity. This gives the knife momentum: momentum=force x time.

My mistake! Yeah gravity too. Ok! So as time passes in air more and more momentum is acquired by body.( I called it force!)

The momentum gained allows it to continue descending despite the added force of friction.

So are the two forces equal at all times and because of momentum it’s going through the cardboard?

But it will lose momentum because the frictional force exceeds the gravitational force, leading to a net upward force.

I don’t get this line.
By the way momentum is next chapter after laws of motion. Why have they given this question here or can it be understood without momentum?
So moving on friction F= -(-mg-mv/t)=-(-2+-2/t)= ?

 

[haruspex]

So are the two forces equal at all times

No. While gravity was unopposed, the knife accelerated, gaining downward momentum. When it hit the cardboard, it encountered a frictional force greater than the force from gravity. The net force now being upwards, that added upward momentum, i.e. it reduced downward momentum. Eventually all the momentum is lost and the knife comes to rest.

can it be understood without momentum?

Yes, but I mentioned it because you wrote that the downward acceleration gave the knife "force". What you were intuitively grappling with was the concept of momentum, which is different from force.

friction F= -(-mg-mv/t)

You will find it less prone to error if you always write such equations in the Newtonian form, (sum of forces) = mass * acceleration, and stick to a sign convention. In this case, the motion being vertical, you should choose to make either upwards positive for all accelerations, forces, velocities.., or downwards positive for all.

 

[rudransh verma]

Yes, but I mentioned it because you wrote that the downward acceleration gave the knife "force". What you were intuitively grappling with was the concept of momentum, which is different from force.

Ok.

While gravity was unopposed, the knife accelerated, gaining downward momentum. When it hit the cardboard, it encountered a frictional force greater than the force from gravity. The net force now being upwards, that added upward momentum, i.e. it reduced downward momentum. Eventually all the momentum is lost and the knife comes to rest.

Ok. Hence the retardation. So downward momentum starts from initial point and is there till all is lost by upward momentum. Net momentum = net force=0 . Body comes to rest.

You will find it less prone to error if you always write such equations in the Newtonian form, (sum of forces) = mass * acceleration, and stick to a sign convention.

true!

 

[jbriggs444]

Ok. Hence the retardation. So downward momentum starts from initial point and is there till all is lost by upward momentum. Net momentum = net force=0 . Body comes to rest.

Be careful. Momentum and force do not have the same units. Saying that momentum is equal to force will never be correct. They are both zero. But they are incommensurable.

 

[haruspex]

Net momentum = net force=0

Since a force cannot equal a momentum, being dimensionally different, it is clearer to write that as momentum =0 (so no velocity), net force=0 (so no acceleration).
Note the parallel : just as acceleration is the rate of change of velocity, force is the rate of change of momentum.

 

[rudransh verma]

Note the parallel : just as acceleration is the rate of change of velocity, force is the rate of change of momentum.

Ok but please solve the problem with or without p.

 

[Steve4Physics]

You may have resolved these questions through what @haruspex has said. But since you asked me directly, here's my reply.

So is it wrong to say “force is also created through acceleration”?

It is wrong. Net force (if non-zero) produces acceleration. Not the other way round.

And do not confuse 'acceleration' (units m/s²) with 'momentum' (units kg.m/s).

No! How? mg is -ve: -mg and force due to falling or acceleration is -5 not 5. So -(-2-5)=7 N (friction force).

The 5N force is the net force as the knife passes through the cardboard. It acts upwards (which is why it slows down the knife) so has a positive value, +5N not -5N.

 

[haruspex]

Ok but please solve the problem with or without p.

You do not need to invoke momentum. Just correct the equations below by including gravity.

So $F=(200*25)/1000$
$F=5 N$

 

[rudransh verma]

You do not need to invoke momentum. Just correct the equations below by including gravity.

the third eqn of motion give acceleration based on the values v, u, s. The question is imagery. It’s mere algebra. But what if it were true in real life. Body falling through cardboard, the distance covered and v and u all combined should have given correct acceleration including g. It looks like while the body is traveling through cardboard, it travels a distance s and the acceleration it undergoes is in the absence of gravity.
Math wise it’s correct but physics wise it’s not, I guess!

 

[jbriggs444]

Math wise it’s correct but physics wise it’s not, I guess!

The math, the physics and the reality all match. It is your understanding of them that is astray.

the third eqn of motion give acceleration based on the values v, u, s. The question is imagery. It’s mere algebra. But what if it were true in real life. Body falling through cardboard, the distance covered and v and u all combined should have given correct acceleration including g.

If what you know is the distance covered $s$, the starting velocity $u$ and the ending velocity $v$ and if you have assumed that the acceleration is constant then you can indeed compute the required acceleration $a$.

One normally accounts for this acceleration by summing the forces and computing a total acceleration: $$F_1 + F_2 = ma$$ $$a = \frac{F_1 + F_2}{m}$$You seem to want to take the forces individually, compute accelerations individually and add the accelerations. $$a = a_1 + a_2 = \frac{F_1}{m} + \frac{F_2}{m}$$So your vision is that the knife is subject to two accelerations which combine to produce the true acceleration that is actually seen?

It looks like while the body is traveling through cardboard, it travels a distance s and the acceleration it undergoes is in the absence of gravity.

It undergoes an acceleration. That acceleration is less than it would have been in the absence of gravity.

 

[rudransh verma]

The math, the physics and the reality all match. It is your understanding of them that is astray.

It matches but after separately adding the of g to a. The eqn of motions should have given correct data. Not like adding the value of g after finding a to get total acceleration.
What is the meaning of third eqn? Third eqn says final velocity of the body is v whose initial velocity is u and it undergoes a acceleration and a displacement s. Relation is $v^2=u^2+2as$.
So it should have given us the final value of acceleration. There should have no need to add g afterwards.
In book they have done this solution which I think is not entirely correct.

 

[jbriggs444]

It matches but after separately adding the value of g to a. The eqn of motions should have given correct data. Not like adding the value of g after finding a to get total acceleration.
What is the meaning of third eqn? Third eqn says final velocity of the body is v whose initial velocity is u and it undergoes a acceleration and a displacement s. Relation is $v^2=u^2+2as$.
So it should have given us the final value of acceleration. There should have no need to add g afterwards.
In book they have done this solution which I think is not entirely correct.

You are not asked for the final value of acceleration. You are asked for the value of friction which would yield this acceleration while the knife is also subject to gravity.

The knife is subject to the force of gravity, so there is a need to model the effect of gravity.

The standard approach to the problem involves adding or subtracting forces. Not adding or subtracting accelerations. [Even though the math works either way]

 

[Steve4Physics]

it travels a distance s and the acceleration it undergoes is in the absence of gravity.

I hope @haruspex will excuse me for adding this...

Gravity always acts on the knife. The knife's weight is always -2N.

The resistance force in the cardboard is +7N.

The net force on the knife in the cardboard is therefore
$F_{net} = -2N +7N = 5N$

The knife in the cardboard behaves the same as if a single +5N force acted on it.

 

[rudransh verma]

[USER=422467]@jbriggs444[/USER] and @Steve4Physics I hope you get it in my post#81. What you are saying guys is also right.

The knife is subject to the force of gravity, so there is a need to model the effect of gravity.

Yeah! But why separately?
This was the original confusion in my OP. Now I guess it’s settled.

 

[jbriggs444]

Yeah! But why separately?

You are asked for friction. You observe the effect of friction and gravity combined. To calculate the effect of friction alone you have to somehow separate the effect of gravity from the effect of friction.

 

[haruspex]

the third eqn of motion give acceleration based on the values v, u, s. The question is imagery. It’s mere algebra. But what if it were true in real life. Body falling through cardboard, the distance covered and v and u all combined should have given correct acceleration including g. It looks like while the body is traveling through cardboard, it travels a distance s and the acceleration it undergoes is in the absence of gravity.
Math wise it’s correct but physics wise it’s not, I guess!

That has nothing to do with what I meant by:

Just correct the equations below by including gravity.

Those equations being:

So F=(200∗25)/1000
F=5N
Wrong answer!

The correction needed is to be clear and consistent about what F represents. Does it represent the net force on the knife or only the force of friction?

 

[vela]

Body falling through cardboard, the distance covered and v and u all combined should have given correct acceleration including g. It looks like while the body is traveling through cardboard, it travels a distance s and the acceleration it undergoes is in the absence of gravity.

It is kind of interesting that multiple people have told you gravity acts on the knife the entire time, but you seem to insist that gravity doesn't act on the knife once it's in the cardboard. Could you elaborate on what your thinking is here?

 

[bob012345]

It is kind of interesting that multiple people have told you gravity acts on the knife the entire time, but you seem to insist that gravity doesn't act on the knife once it's in the cardboard. Could you elaborate on what your thinking is here?

I wonder if it is how a student might interpret a problem if the book refers to an equation without $g$ explicitly in it as meaning gravity is not involved. But the OP has since seemed to understand that gravity is involved through the whole problem and this discussion is going in circles.

 

[rudransh verma]

Everyone, I am silly! My mistake . The eqn of motion has given the correct value of acceleration. That is net acceleration(25). And in the back of my head I was thinking it’s not. I cannot track everything where I went wrong but you guys understand. By the way I was not switching off the gravity. Might have by Mistake somewhere!
$$Net force= friction-mg$$
$$ma= ma’-mg$$
$$a’=a+g$$
$$a’=25+10$$
$$a’=35 m/s^2$$
So,$$friction= .2*35=7 N$$

 

[jbriggs444]

Everyone, I am silly! My mistake . The eqn of motion has given the correct value of acceleration. That is net acceleration(25). And in the back of my head I was thinking it’s not. I cannot track everything where I went wrong but you guys understand. By the way I was not switching off the gravity. Might have by Mistake somewhere!
$$Net force= friction-mg$$
$$ma= ma’-mg$$
$$a’=a+g$$
$$a’=25+10$$
$$a’=35 m/s^2$$
So,$$friction= .2*35=7 N$$

Why are you dividing by $m$ only to end up multiplying by $m$?

 

[rudransh verma]

Why are you dividing by m only to end up multiplying by m?

 

[Steve4Physics]

Knowing what you now know, what would your working be if this were a piece of homework or an examination question?

 

[rudransh verma]

Knowing what you now know, what would your working be if this were a piece of homework or an examination question?

Simple , upto a= 25 m/s^2 is right. Then just add the above part. Question done! 200 marks out of 100.

 

[kuruman]

Simple , upto a= 25 m/s^2 is right. Then just add the above part. Question done! 200 marks out of 100.

It's not that simple for me because I am having trouble sorting out the logical sequence "up to a = 25 m/s^2" in all these posts (93 and counting). Would it be possible for you to post the entire solution, from beginning to end, in one complete post? Thank you.

 

[Steve4Physics]

Simple , upto a= 25 m/s^2 is right. Then just add the above part. Question done! 200 marks out of 100.

Your working up to finding a=25m/s² did not correctly apply a stated sign-convention. Purely by chance, you ended up with the correct sign (+25m/s², assuming we are using upwards-is-positive).

What you call the 'above part' (assumed to mean your Post #89 working) shows a confused approach and will not earn all the 'method' marks.

As a (retired) physics teacher assessing a piece of homework, I award 6/10.

 

[Steve4Physics]

@rudransh verma, sorry to see the sad face. Maybe I was a bit harsh.

Do you understand why you didn't get 10/10?

To paraphrase George Santayana somewhat: if we don't learn from our mistakes, we are destined to repeat them.

 

[rudransh verma]

$$v^2=u^2+2as$$
$$v=+-10m/s$$
Taking -ve sign!
Again $$v^2=u^2+2as$$
$$a=+25m/s^2$$

So, $$Friction= F net+mg$$
$$(.2*25 + (.2)10)$$
$$Friction= +7 N$$

 

[Steve4Physics]

$$v^2=u^2+2as$$
$$v=+-10m/s$$
Taking -ve sign!
Again $$v^2=u^2+2as$$
$$a=+25m/s^2$$

So, $$Friction= F net+mg$$
$$(.2*25 + (.2)10)$$
$$Friction= +7 N$$

That’s much better. 9/10 (or 10/10 on a good day)!

But the easy/quick method (hinted at in earlier posts) is to forget about velocities, accelerations and forces. Using gravitational potential energy and work, the problem can easily be solved in about 3 lines. Can you do it that way?

 

[kuruman]

$$v^2=u^2+2as$$ $$v=+-10m/s$$ Taking -ve sign!
Again $$v^2=u^2+2as$$ $$a=+25m/s^2$$ So, $$Friction= F net+mg$$ $$(.2*25 + (.2)10)$$ $$Friction= +7 N$$

Thank you for the detailed solution. I think that you are unclear about the meaning of the symbols in your first equation. I understand that you have taken "up" as positive and "down" as negative. I also understand that in the expression $$v^2=u^2+2as$$ you have taken $v =-10$ m/s to be the velocity of the knife just before it hits the cardboard and $u=0$ is the final velocity of the knife. That is all correct and consistent. My question is, what about $s$? It should represent the displacement of the knife. Since the knife ends up below its initial position, the displacement must be negative, i.e. $s=-2~$m. If I substitute, I get $$(-10~\text{m/s})^2=0^2+2a(-2\text{m})$$ $$100~(\text{m/s})^2=-4~a~(\text{m})$$ $$a=-25~\text{m/}{\text{s}}^2.$$Something isn't right because we know that the acceleration is positive.

I am not trying to confuse you but to show you that there is still some confusion in your head about the equation $v^2=u^2+2as$, namely how you use it and what the symbols mean. Thinking of $v$ as the initial velocity and $u$ as the final velocity can lead you into trouble as it has here. A better way to write the same equation is to use subscripts to match points in space with velocities at these points: $$2a(s_2-s_1)=v_2^2-v_1^2.$$Then, if we assume that the edge of the cardboard where the knife hits is at zero height, we have $s_1=0$, $v_1=-10~\text{m/s}$, $s_2=-2~\text{m}$, $v_2=0$ so that $$2a(-2~\text{m}-0)=0^2-(-10~\text{m/s})^2$$ $$-4~a~(\text{m})=-100~(\text{m/s})^2$$ $$a=+25~\text{m/}{\text{s}}^2.$$Whether you continue using your equation or adopt the one I suggested is up to you. To quote Confucius, "If you make a mistake and do not correct it, this is called a mistake."

 

[Steve4Physics]

the displacement must be negative, i.e. $s=-2~$m. If I substitute, I get $$(-10~\text{m/s})^2=0^2+2a(-2\text{m})$$ $$100~(\text{m/s})^2=-4~a~(\text{m})$$ $$a=-25~\text{m/}{\text{s}}^2.$$Something isn't right because we know that the acceleration is positive.

I think @rudransh verma's working (in full) is
$v_f^2 = v_i^2+ 2as$ we get
$0^2 = (-10)^2 + 2a(-2)$
$4a = 100$
$a = 25m/s$ (positive as required)

 

[kuruman]

I think @rudransh verma's working (in full) is
$v_f^2 = v_i^2+ 2as$ we get
$0^2 = (-10)^2 + 2a(-2)$
$4a = 100$
$a = 25m/s$ (positive as required)

I am not sure. @rudransh verma uses $v$ on the left and $u$ on the right side of his equation without subscripts that specify what they stand for. Furthermore, he clearly identifies v = ±10 m/s. Why did you feel the need to use subscripts $i$ and $f$ in your equation which he doesn't use? My point is exactly yours: without subscripts one can get easily confused about what's what and where. I think the most efficient way to write this equation is in the form of the work-energy theorem per unit mass $$\vec a \cdot \vec s=\frac{1}{2}\left(v_{\!f}^2-v_{\!i}^2 \right)$$ which makes it independent of one's choice of axes.

 

[Steve4Physics]

Sighs and takes deep breath.

@rudransh verma has omitted some detailed steps, which makes properly checking his working impossible. On reflection, I think you (@kuruman) may well be right and he (guessing ‘he’) hasn’t solved the problem correctly.

I am not sure. @rudransh verma uses $v$ on the left and $u$ on the right side of his equation without subscripts that specify what they stand for. Furthermore, he clearly identifies v = ±10 m/s.

I've assumed the standard equation $v^2=u^2+2as$ has been used correctly. Maybe I'm too trusting!

He (assuming male) goes on and chooses v=-10m/s as the final velocity after the free-fall, which is correct.

Why did you feel the need to use subscripts $i$ and $f$ in your equation which he doesn't use?

For disambiguation.

In Post #99 you wrote:
“you have taken $v=-10$m/s to be the velocity of the knife just before it hits the cardboard and $u=0$ is the final velocity of the knife. That is all correct and consistent “

I disagreed.

To be 'correct and consistent' requires that initial velocity $u=-10$m/s and final velocity $v=0$. Using the standard equation $v^2=u^2+2as$ for the 2m drop then correctly gives $a=+25m/s^2$.

I posted my version of this part of the calculation, using $v_i$ and $v_f$ for the 2m drop to avoid confusion with u and v from the free-fall drop.

But things are getting convoluted and it's getting too hard to explain clearly.

My point is exactly yours: without subscripts one can get easily confused about what's what and where. I think the most efficient way to write this equation is in the form of the work-energy theorem per unit mass $$\vec a \cdot \vec s=\frac{1}{2}\left(v_{\!f}^2-v_{\!i}^2 \right)$$ which makes it independent of one's choice of axes.

I agree suffixes should be used to avoid ambiguities.

My overall preferred approach to the whole problem is to consider the loss of gravitational potential energy and the work done by the resistive force. The question is then easily solved with only 2 or 3 lines of trivial working.

 

[kuruman]

To @Steve4Physics: You don't have to explain in such detail - I am sure you know what you're doing. Actually, the question is even more easily solved when one applies the relevant equation piecewise to the overall motion from start to finish in which case $v_i=v_f=0.$ Then $$a_1s_1 +a_2s_2=0$$ where $a_1=-g$ and $a_2=-g+a$. Solving this for $a$ and multiplying by the mass gives the answer.

 

[Steve4Physics]

Actually, the question is even more easily solved when one applies the relevant equation piecewise to the overall motion from start to finish in which case $v_i=v_f=0.$ Then $$a_1s_1 +a_2s_2=0$$ ...

That's a really neat trick I've not seen before.

It works whenever $v_i=v_f$ (not only when they are both zero). And it can be extended to any number of 'pieces'.

The equivalent time version could also be useful in the toolkit:
If $v_i=v_f$ then $a_1t_1 +a_2t_2 + ...=0$.