Conservation of momentum and energy problem-Two particles conected by string

[Tanya Sharma]

Homework Statement

Two particles A and B each of mass m are attached by a light inextensible string of length 2l .The whole system lies on a smooth horizontal table with B initially at a distance l from A.The particle at end B is projected across the table with speed u perpendicular to AB.Find the velocity of ball A just after the jerk ?

Homework Equations

The Attempt at a Solution

The particle at A feels jerk when the string becomes taut.When this happens the angle θ wiill be 30° .

V=velocity of A after the jerk
$u_x$ = x component of velocity of B after the jerk
$u_y$ = y component of velocity of B after the jerk

Using conservation of momentum in y direction we have $u_y$=vsin30° =v/2

Using conservation of momentum in x direction we have mu = mvcos30°+m$u_x$

Applying conservation of energy we have

$\frac{1}{2}mu^2 = \frac{1}{2}mv^2+\frac{1}{2}m{u_x}^2+\frac{1}{2}m{u_y}^2$

Putting the values of $u_x$ and $u_y$ in the above equation and solving
we get v=$\frac{\sqrt{3}}{2}u$

This is not the correct answer ...Please guide me where am i getting it wrong .

 

[tiny-tim]

Hi Tanya!

Applying conservation of energy we have …

This is a collision.

Energy is never conserved in a collision unless the question says so!

Hint: you obviously need some other equation … in which direction do you think A will move? ​

 

[Tanya Sharma]

Hello Sir...thanks for the response ...but this is not a collision
B is given a velocity horizontal as given in the diagram...b will move
along with the string and when the string becomes taut, then particle A
will feel a jerk ...This velocity of A is what we have to find ...since no external forces
act hence we can conserve the momentum ...Since there are no dissipative
forces hence we are conserving the energy .

 

[tiny-tim]

Hi Tanya!

… this is not a collision
…
Since there are no dissipative forces hence we are conserving the energy .

Yes it is, and no we're not!

Energy is usually only conserved in smooth processes. This is a jerk.

For example, a block sliding down a ramp onto the ground may have energy conservation if the ramp curves to meet the ground smoothly, but not if there's an angle.

(oh, and don't call me sir … I'm only a little goldfish! )

 

[Tanya Sharma]

Okay...then please help me how should i approach the problem...B will move in horizontal direction and when the string becomes taut ...A will feel the tension and will move in the direction of string joining A and B ..isnt it??

How is the energy dissipated in this problem?

 

[tiny-tim]

A will feel the tension and will move in the direction of string joining A and B ..isnt it??

That's correct, the only force on A is from the string, so it must be along the string (because it's a string!), so the acceleration will be along the string.

(Same for B … the acceleration will be along the string)

How is the energy dissipated in this problem?

sound, heat, and vibration

 

[Tanya Sharma]

Lets focus on B's motion ...I agree that the acceleration of B will be along the string ...Now let's resolve this accleration in two components . One along the y direction and the other along the x direction...The component along the x direction will decelerate B and the one along y direction being normal will change direction of B. Am i right?

But then A and B needs to have same velocity along the string .Isnt it??

 

[tiny-tim]

Now let's resolve this accleration in two components .

it might be easier to use the string as one of the component directions

The component along the x direction will decelerate B and the one along y direction being normal will change direction of B. Am i right?

not really following this

mass times acceleration of A will equal minus mass times acceleration of B

But then A and B needs to have same velocity along the string .Isnt it??

no, there could be rotation

 

[Tanya Sharma]

The component along the x direction will decelerate B and the one along y direction being normal will change direction of B. Kindly see the attachment.The tension will act along the string .This tension will have two components ,one along x and other along y direction.The one along y direction will be normal so will change only direction .The other along x direction will decelarate B .


tiny tim...i am absolutely confused...I have tried hard on the problem...Kindly explain the motion of the two particles in this particular problem...
I understand that accelaration will be along the string but what about the velocity of B ?

 

[tiny-tim]

Two particles A and B each of mass m are attached by a light inextensible string of length 2l .The whole system lies on a smooth horizontal table with B initially at a distance l from A.The particle at end B is projected across the table with speed u perpendicular to AB.Find the velocity of ball A just after the jerk ?

A and B continue at distance 2l apart.

The string joining them will rotate, but will remain taut (of length 2l).

So d/dt (r_A - r_B).(r_A - r_B) = 0, ie (v_A - v_B).(r_A - r_B) = 0.

 

[Tanya Sharma]

Please have a look at Post #9...i have edited it

 

[ehild]

Instead of conservation of energy, the angular momentum is conserved in addition to conservation of linear momentum. Write up the equations for the x, y components of momentum and the angular momentum before the jerk and after it.
You have one more equation which follows from the constant length of the string and Tiny-Tim suggested to you: (v_A - v_B).(r_A - r_B) = 0

ehild

 

[Tanya Sharma]

ehild... thank you very much for the response

You have one more equation which follows from the constant length of the string and Tiny-Tim suggested to you: (v_A - v_B).(r_A - r_B) = 0

ehild

how did we get this equation and what does it mean?

and why are we conserving angular momentum here ... i mean how can we say that the system is rotating?

 

[ehild]

With respect to point A the mass B has angular momentum: How is it defined?

The distance between B and A is constant after the string gets taut. It is the magnitude of the difference r_B-r_A:
(r_B-r_A)²=4l². Taking the time derivative you get the equation (r_B-r_A)(v_B-v_A)=0.

ehild

 

[Tanya Sharma]

m(vB - vA).(rB - rA) is the angular momentum of B w.r.t A

 

[ehild]

m(vB - vA).(rB - rA) is the angular momentum of B w.r.t A

It is the difference between the angular momenta of B and A. Write the angular momentum of the whole system with respect to a fix point, say A. Just after the jerk, the mass at A is sill at its original place.

ehild

 

[Tanya Sharma]

Okay...What will be the motion of B after the jerk? Will it continue to move forward or will it move along the string length ?

 

[ehild]

Okay...What will be the motion of B after the jerk? Will it continue to move forward or will it move along the string length ?

Both masses will move, in such way that the distance is constant between them. You can figure out the velocity of both masses from the conservation equations.

ehild

 

[Tanya Sharma]

tiny tim and ehild...thank you very much

 

[ehild]

and why are we conserving angular momentum here ... i mean how can we say that the system is rotating?

The angular momentum is the vector product of the position vector and the linear momentum: m[rxV]. If an object moves along a straight line which does not go through the origin it has angular momentum with respect to the origin. No need to travel along a circle.

You hang a small object with a string to a peg. Hit the hanging
mass with a horizontal force: It gets a horizontal velocity and it will move along a circle won't it? It gains angular momentum from the torque applied.
Put a stick on the table and hit one end: the stick will move in a direction and it will also turn. It got some linear momentum from the force applied for a short time, and also angular momentum from the applied torque.

Mass B has some nonzero angular momentum with respect to the original position of body A. We choose that point as origin. The system of the bodies and the string connecting them is a closed system: there are neither external forces nor external torques. Both the linear momentum and the angular momentum of the whole system is conserved.

You do not know what exactly happens when the string gets taut. There is a big tension evolved in the string for a very short time, and that tension accelerates both bodies at the ends. The tension acts along the string, so the accelerations happen in the direction of the taut string. The string is said unextendable, but that means only that the extension is very small while the tension is very high. Moreover, the process happening in the string can be quite complicated. It gets stretched, the molecules move out from the equilibrium, and colliding with the others, some of the elastic energy transforms to heat. That is why the energy is not conserved. At the end, the bodies reach their new velocities with equal components along the string, the tension and the (very small) strain in the string becomes constant. The centre of mass moves with constant velocity and the bodies rotate around the CM. The whole system behaves like a rigid body. But the in-plane general motion of a rigid body means its CM traveling and also rotation about the CM.

ehild

 

[tiny-tim]

… why is energy conserved in the problem https://www.physicsforums.com/showthread.php?t=642703 but not in this problem?

Three identical balls are connected by light inextensible strings with each other as shown and rest over a smooth horizontal table .At the moment t=0,ball B is imparted a velocity.Calculate the velocity of A when it collides with ball C.

it that problem, you are given the (initial) velocity after the jerk, but in this problem, you are given the velocity before the jerk

energy is never conserved during a jerk, but it should be conserved in the smooth motion that follows the jerk

 

[Tanya Sharma]

tiny tim and ehild ...I have attempted the problem ...Kindly check my work

Consider A as the origin and +x axis to be rightwards and +y axis to be downwards

u = initial velocity of B rightwards
V = velocity of A after the jerk

$W_x$ = x component of velocity of B after the jerk
$W_y$ = y component of velocity of B after the jerk
$r_A$ = position vector of A
$r_B$ = position vector of B

Applying conservation of momentum in x direction we have

$mu=mVcos30° + mW_x$
$W_x=u-\frac{\sqrt{3}}{2}V$ (1)

Applying conservation of momentum in y direction we have

$0 = mVsin30°+mW_y$
$W_y=-V/2$ (2)


$V_A = Vcos30°\hat{i}+Vsin30°\hat{j}$
$V_B=W_x\hat{i}+W_y\hat{j}$

$r_A=0$
$r_B=\sqrt{3}\hat{i}+\hat{j}$

Since string length remains constant we have $(V_A-V_B).(r_A-r_B) = 0$

$[(Vcos30°-W_x)\hat{i}+(Vsin30°-W_y)\hat{j}].[\sqrt{3}\hat{i}+\hat{j}] = 0$

$\sqrt{3}(Vcos30°-W_x) + (Vsin30°-W_y) = 0$

Putting the values of $W_x$ and$W_y$ from eq 1 and 2 we get

$\sqrt{3}[\frac{\sqrt{3}}{2}V-(u-\frac{\sqrt{3}}{2}V)] + (V/2+V/2) = 0$

$\sqrt{3}(\sqrt{3}V-u) + V = 0$

$4V-\sqrt{3}u = 0$

$V=\frac{\sqrt{3}}{4}u$

 

[ehild]

Tania,
Although your solution corresponds to the mirror image of the figure in the first post, it looks correct
You supposed that particle A gains velocity in the direction of the string after the jerk. That looks straightforward, but I would not be so sure in it. Anyway, I got the same result as you assuming conservation of both momentum and angular momentum. You need only a little correction: The velocity of particle A was the question, and you gave the speed.
Anyway, it is a nice work, congratulation!


ehild