Find in terms of ##h## when ##P## comes instantaneously to rest

[SammyS]

That was my own. I was telling them how I started the problem, instead of just writing down random equations.

If you are requesting to see my diagram I’m on my phone, I don’t have it with me currently.

OK. All I saw was the one posted by @chwala in Post #16.

@erobz : You have done an excellent job with Thread, showing great patience.

 

[erobz]

OK. All I saw was the one posted by @chwala in Post #16.

@erobz : You have done an excellent job with Thread, showing great patience.

Thanks @SammyS(that means a lot), sometimes I can hold it together…

 

[Steve4Physics]

@erobz, I concur with @SammyS's compliment!

@chwala, it may be worth noting that the question can be quickly/easily solved using conservation of energy. Use $\Delta (KE) + \Delta(GPE) = 0$. There is no need to find the tension and acceleration.

Try it for yourself first. Here’s my version:

Spoiler: Energy solution

Between release and Q hitting the ground, Q drops a distance $h$ so P rises a distance $h$. P is now a distance $2h+h=3h$ above the ground. Both P and Q are then travelling with the same speed, $v$.

Applying conservation of energy, with a little care we can immediately write:

$\frac 12(2m)v^2 +\frac 12 (5m)v^2 + (2m)gh + (5m)g(-h) = 0$

A little simple algebra gives $v^2 = = \frac {6gh}7$

The string is now slack so P becomes a vertical projectile with initial vertical velocity $v$, rising a further distance $S$ (from a height of $3h$). Applying conservation of energy to P gives:

$-\frac 12(2m)v^2 + (2m)gS = 0$ (or we could just use a simple kinematics equation)

Substituting $v^2 = \frac {6gh}7$ in the above equation and a little simple algebra gives $S =\frac 37 h$.

So P’s final height is $3h + \frac 37 h = \frac {24}7 h$.

 

[erobz]

@erobz, I concur with @SammyS's compliment!

Thank You!

 

[chwala]

@erobz, I concur with @SammyS's compliment!

@chwala, it may be worth noting that the question can be quickly/easily solved using conservation of energy. Use $\Delta (KE) + \Delta(GPE) = 0$. There is no need to find the tension and acceleration.

Try it for yourself first. Here’s my version:

Spoiler: Energy solution

Between release and Q hitting the ground, Q drops a distance $h$ so P rises a distance $h$. P is now a distance $2h+h=3h$ above the ground. Both P and Q are then travelling with the same speed, $v$.

Applying conservation of energy, with a little care we can immediately write:

$\frac 12(2m)v^2 +\frac 12 (5m)v^2 + (2m)gh + (5m)g(-h) = 0$

A little simple algebra gives $v^2 = = \frac {6gh}7$

The string is now slack so P becomes a vertical projectile with initial vertical velocity $v$, rising a further distance $S$ (from a height of $3h$). Applying conservation of energy to P gives:

$-\frac 12(2m)v^2 + (2m)gS = 0$ (or we could just use a simple kinematics equation)

Substituting $v^2 = \frac {6gh}7$ in the above equation and a little simple algebra gives $S =\frac 37 h$.

So P’s final height is $3h + \frac 37 h = \frac {24}7 h$.

This looks nice! interestingly, the MS did not envisage this approach. Does a student get full marks by using this approach? This question is from As level Mechanics Paper.

 

[Steve4Physics]

This looks nice! interestingly, the MS did not envisage this approach. Does a student get full marks by using this approach? This question is from As level Mechanics Paper.

You get full marks in A-level (and no doubt other) examinations for a correct alternative approach. (I write as someone who has worked for examination boards, marking A-level papers, in the past.)

 

[chwala]

now my problem is here; when the rope is slack, after $Q$ hits the ground we have particle $P$ travelling the extra distance given by the equation,

$v^2 =u^2+2(-g)h$ and noting that $u=\sqrt{8.4h}$,

$0 = 8.4h -2gS$

$0 = \dfrac{6gh}{7} -2gS$

$2gS=\dfrac{6gh}{7}$

$2S=\dfrac{6h}{7}$

$S=\dfrac{3h}{7}$

Aaaaaaaaaaaah! I can see it now! The distance from ground will now be given by $2h + h + S= 2h + h +\dfrac{3h}{7}$

I just re-visted this, now clear,

Velocity of P upwards = Velocity of Q downwards = $\sqrt {\dfrac{6gh}{7}}$. The distance travelled by P here is $h$.
Secondly, When Q stops, P continues to move with $u=\sqrt {\dfrac{6gh}{7}}$, and on using $v^2=u^2-2gs$
$0=\dfrac{6gh}{7}-2gs$
$s=\dfrac{3h}{7}$.
Therefore, P travels a total of , $\left[2h+h+\dfrac{3h}{7}= \dfrac{24h}{7}\right]$.