# Math Help - Help needed in Mechanics 1- Potential energy!

1. ## Help needed in Mechanics 1- Potential energy!

Guys m stuck over these... help needed!

Q1. Particles of mass 1.2 kg and 1.4 kg hang at the same level, connected by a long light inextensible string passing over a small smooth peg. They are released from rest with the string taut. Calculate the separation of the particles when they are moving with speed 0.5 metres per second.
Ans: 0.325 m.

Q2. A particle of mass 2.2 kg rests on a smooth slope inclined at 30 degrees to the horizontal. It is connected by a light inextensible string passing over a smooth rail at the top of the slope to a particle of mass 2.7 kg which hangs freely. The particles are set in motion by projecting the lighter down a line of greatest slope with 4 metres per sec. Find the distance the particles travel before their direction of motion is reversed. Find the total energy gained by the hanging mass during this part of the motion, and hence find the tension in the string.
Ans.2.45m , 44.55 J, 18.2 N.

I am completely stuck at 1. However for the second question i am getting -2.45 m so there's probably some mistake in my approach to the question. And kindly guys, try to solve these with the help of energy conservation principle.

Thanks!

2. ## Re: Help needed in Mechanics 1- Potential energy!

(1) for simplicity, let the masses start their motion from rest at the same position ... mark this initial position as the zero for GPE.

total mechanical energy of the system = 0

after release, large mass falls a distance x ... its energy is $\frac{1}{2}Mv^2 - Mgx$

small mass rises the same distance x ... its energy is $\frac{1}{2}mv^2 + mgx$

sum of both energies = 0

$\frac{1}{2}Mv^2 - Mgx + \frac{1}{2}mv^2 + mgx = 0$

$\frac{1}{2}(M+m)v^2 = gx(M-m)$

$\frac{(M+m)v^2}{g(M-m)} = x$

Thank u

4. ## Re: Help needed in Mechanics 1- Potential energy!

They have taken g=10 to get x=0.325 So isnt the separation 2x=0.650?

5. ## Re: Help needed in Mechanics 1- Potential energy!

Let A be the 2.2 kg mass and B be the 2.7 kg mass Let x=required distance. I notice you are taking g=10
A loses PE 2.2*10*xsin30=11x A losesKE (1/2)*2.2*16=17.6 B loses KE (1/2)*2.7*16=21.6 So total loss of energy= 11x+39.2
Gain of energy is the PE of B =2.7*10*x=27x
So 27x=11x+39.2 16x=39.2 x= 2.45
B gains energy 27x-21.6 =44.55
Gain in energy of B = work done by tension T So Tx= 44.55 So T= 44.55/2.45= 18.2

6. ## Re: Help needed in Mechanics 1- Potential energy!

Originally Posted by biffboy
They have taken g=10 to get x=0.325 So isnt the separation 2x=0.650?

I still dont get it...

7. ## Re: Help needed in Mechanics 1- Potential energy!

Here's a different approach. (Nice solution BTW biffboy).
For the larger mass let down be the positive direction and suppose that the tension in the string is $T.$
Then, (F=ma),
for the larger mass
$2.7g-T=2.7a.................(1),$
and for the mass on the slope,
$T-2.2g\cos60=2.2a........(2).$
Adding, (and taking $g$ to be $10),$
$27-11=4.9a,$ so $a=16/4.9$
Now use one of the equations of motion for a body moving with constant acceleration $v^{2}=u^{2}+2as,$
and we have, when the large mass comes to rest,
$0=16+2(16/4.9)s,$ so $s=-16*4.9/(2*16)=-2.45$
(Since down is the positive direction, the negative sign reflects the fact that the larger mass has moved upwards before coming (momentarily) to rest.)
The gain in energy of the larger mass will be its gain in PE minus its loss in KE
$=2.7g*2.45-(1/2)2.7*4^{2}=44.55$
The tension in the string we can get from (1) or (2),
From (1), $T=2.7g-2.7a=27-2.7(16/4.9)=18.18$

8. ## Re: Help needed in Mechanics 1- Potential energy!

That's the way I would have done it left to myself. I used energy because of the title of the question.

9. ## Re: Help needed in Mechanics 1- Potential energy!

Originally Posted by BobP
Here's a different approach. (Nice solution BTW biffboy).
For the larger mass let down be the positive direction and suppose that the tension in the string is $T.$
Then, (F=ma),
for the larger mass
$2.7g-T=2.7a.................(1),$
and for the mass on the slope,
$T-2.2g\cos60=2.2a........(2).$
Adding, (and taking $g$ to be $10),$
$27-11=4.9a,$ so $a=16/4.9$
Now use one of the equations of motion for a body moving with constant acceleration $v^{2}=u^{2}+2as,$
and we have, when the large mass comes to rest,
$0=16+2(16/4.9)s,$ so $s=-16*4.9/(2*16)=-2.45$
(Since down is the positive direction, the negative sign reflects the fact that the larger mass has moved upwards before coming (momentarily) to rest.)
The gain in energy of the larger mass will be its gain in PE minus its loss in KE
$=2.7g*2.45-(1/2)2.7*4^{2}=44.55$
The tension in the string we can get from (1) or (2),
From (1), $T=2.7g-2.7a=27-2.7(16/4.9)=18.18$
Thanks...but i have already solved it through this approach... The topic m doing now a days involves energy equations so i just needed some guidance about this question with reference to energies and stuff.... Thanks a lot though, u helped clear my confusion regarding the minus sign!!!