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Hey guys,

I've been trying to revise the connected particles chapter and i am getting nowhere!.. I'm just so confused about everything in this part of M1. Anybody got any tips or any help they could give. I have a question that i have been told if you can do, you do not need 2 worry about any connected particles questions because it has everything in there you need 2 solve. If anybody could answer it step-by-step for i would appreciate greatly.

A particle A of mass 0.8kg rests on a horizontal table and is attached to one end of a light inextensible string. The string passes over a small pulley P fixed at the edge of the table. The other end of the string is attached to a particle B of mass 1.2kg which hangs freely below the pulley. The system is released from rest with the string taut and with B at a height of 0.6m above the ground. In the subsequent motion A does not reach P before B reaches the ground. In an inital model of the situation, the table is assumed to be smooth.

(a) Find the tension in the string before B reaches the ground

(b) The time taken for B to reach the ground

In a refinement of the model, it is assumed that the table is rough and that the coefficient of friction between A and the table is 1/5.

(c) Find the time taken for B to hit the ground.

I know this is a lot to ask but any help is very much appreciated.

Thanx

Tut

I've been trying to revise the connected particles chapter and i am getting nowhere!.. I'm just so confused about everything in this part of M1. Anybody got any tips or any help they could give. I have a question that i have been told if you can do, you do not need 2 worry about any connected particles questions because it has everything in there you need 2 solve. If anybody could answer it step-by-step for i would appreciate greatly.

A particle A of mass 0.8kg rests on a horizontal table and is attached to one end of a light inextensible string. The string passes over a small pulley P fixed at the edge of the table. The other end of the string is attached to a particle B of mass 1.2kg which hangs freely below the pulley. The system is released from rest with the string taut and with B at a height of 0.6m above the ground. In the subsequent motion A does not reach P before B reaches the ground. In an inital model of the situation, the table is assumed to be smooth.

(a) Find the tension in the string before B reaches the ground

(b) The time taken for B to reach the ground

In a refinement of the model, it is assumed that the table is rough and that the coefficient of friction between A and the table is 1/5.

(c) Find the time taken for B to hit the ground.

I know this is a lot to ask but any help is very much appreciated.

Thanx

Tut

1. Consider f = ma for both particles. Remember that the only force accelerating this system, initially, is the 1.2g which B will have, due to its weight.

So, for A, you get T = 0.8a

and for B, you get 1.2g - T = 1.2a

(Note the order of the T and the g. The tension is the resistive force for B and the main pulling force for the object A, and assuming that the system moves downwards you have to have them in this order.)

Now simply add the two equations, T cancels. 1.2g = 2a, a => 3/5 g.

Subbing back into either of the equations, T = 0.8(3/5 g)

Solve for T. T => 12/25 g. (or 0.48g)

2. Using s = ut + (1/2)at^2. Simply let u = 0, a (as we already know) = 3/5 g. and s = 0.6. And solve for t. t => 0.45.

3. We just need to reconsider the situation for B now. Taking into account the fact that its now T - µ(0.8g) = 0.8a.

F = µR, F = (1/5) * (0.8g) = 4/25 g

T - (4/25)g = 0.8a (new equation for A)

1.2g - T = 1.2a

Again, just add to eliminate T.

1.04g = 2a

a => 0.52g.

Finally, using s = ut + (1/2)at^2 again, using u = 0, s = 0.6m and a = 0.52 g.

Solve to find that t => 0.49s.

So, for A, you get T = 0.8a

and for B, you get 1.2g - T = 1.2a

(Note the order of the T and the g. The tension is the resistive force for B and the main pulling force for the object A, and assuming that the system moves downwards you have to have them in this order.)

Now simply add the two equations, T cancels. 1.2g = 2a, a => 3/5 g.

Subbing back into either of the equations, T = 0.8(3/5 g)

Solve for T. T => 12/25 g. (or 0.48g)

2. Using s = ut + (1/2)at^2. Simply let u = 0, a (as we already know) = 3/5 g. and s = 0.6. And solve for t. t => 0.45.

3. We just need to reconsider the situation for B now. Taking into account the fact that its now T - µ(0.8g) = 0.8a.

F = µR, F = (1/5) * (0.8g) = 4/25 g

T - (4/25)g = 0.8a (new equation for A)

1.2g - T = 1.2a

Again, just add to eliminate T.

1.04g = 2a

a => 0.52g.

Finally, using s = ut + (1/2)at^2 again, using u = 0, s = 0.6m and a = 0.52 g.

Solve to find that t => 0.49s.

a box A of mass 0.8kg rest on a rough horizontal table and is attached to one end of a light inextensible string. the string passes over a smooth pulley fixed at the edge of the table. The other end of the string is attached to a sphere B of the edge of the table. The other end of the string is attached to a sphere B of mass 1.2kg, which hangs freely below the pulley. The magnitude of the frictional force between A and the table is FN. The system is released from rest with the string taut. After release, B DESCENDS A DISTANCE OF 0.9M IN 0.8S . Modelling A and B as particles, calculate: (a) the acceleration of B (b) the tension in the string (c) the value of F (d) find the total distance travelled by A

Original post by Simply_The_best

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