Can You Solve This Motion Problem Using a Simple Trick?

[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 $a_1t_1 +a_2t_2 + ...=0$.