Can You Solve This Motion Problem Using a Simple Trick?

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