Discussion about math, puzzles, games and fun. Useful symbols: ÷ × ½ √ ∞ ≠ ≤ ≥ ≈ ⇒ ± ∈ Δ θ ∴ ∑ ∫ π -¹ ² ³ °

You are not logged in.

- Topics: Active | Unanswered

Pages: **1**

**jimi70****Member**- Registered: 2012-04-04
- Posts: 22

'Weights of 3 kg and 4 kg are connected by a light inextensible string, which passes over a smooth peg. The system is released from rest. Find the acceleration of the system.

After two seconds, the heavier mass strikes the floor and comes to an instantaneous halt. Find the time before the string is again taut.'

I've worked out the acceleration to be 1.4 m/s/s but I'm not sure about the second part. Can somebody help please?

Offline

**bob bundy****Moderator**- Registered: 2010-06-20
- Posts: 7,016

hi jimi70

I get 1.4 as well.

Once the 4Kg has hit the ground it will stay there a while. It is no longer providing any tension to continue the acceleration of the 3 Kg mass so that one will just go up under gravity with a certain start velocity until it slows to a stop and comes back down again. When it gets back to the same place again, the string will go taut (so the 3 Kg can impart an impulse that starts the 4 Kg off again.)

So use the info to work out the common velocity as the 4 Kg hits the ground. Using that as a start velocity then work out how long the 3 Kg will go upwards under gravity before it slows to a stop and comes back down. The string will remain slack until it gets back to the same point. Double this time, as it will take the same time to return to where it was.

[This all assumes that the string is long enough so that the 3 Kg doesn't flip over the peg and land on top of the other mass. The questioner should really have said this won't happen so you could get 'smug points' for pointing it out. ]

Bob

You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei

Offline

**Mrwhy****Member**- Registered: 2012-07-02
- Posts: 52

If the string does not stretch it might help to use energy.

For when one weight hits the floor all its kinetic energy is lost, leaving just some of the total previous enrgy for the other weight (potential and kinetic)

Offline

**jimi****Guest**

I get the following:

The 4 kg weight's displacement is 2.8 m in 2 s. Then the 3 kg weight moves under gravity (taken to be 9.8 m/s/s) alone and, using s = ut + at^2/2, t = (4 + 8 sqrt 2)/14, i.e. 1.09 s. However, my book does not agree. Am I doing something wrong?

**bob bundy****Moderator**- Registered: 2010-06-20
- Posts: 7,016

hi jimi

Not what I'm getting either. I think it is 4/7 sec.

Why are you using s = ...... ? That's for distance. Use v = u + at.

Bob

You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei

Offline

**jimi70****Member**- Registered: 2012-04-04
- Posts: 22

bob,

Can you tell me what is the condition for when the string goes taut?

Offline

**bob bundy****Moderator**- Registered: 2010-06-20
- Posts: 7,016

hi jimi70

The heavier weight (W) is dragging the lighter one (L). When W hits the floor, L is going up with the same velocity. W cannot go any further but L can, so the string goes slack. Let's call that point for L, the point X.

Now, L can go up for a while under gravity; it slows, stops and starts back down. All this time the string is slack. But when L gets back to point X the string is back to where it was. L is going down; the distance from L measured up to the peg and down again towards W is equal to the length of the string. So, at that moment the string is taut again.

Bob

You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei

Offline

**jimi70****Member**- Registered: 2012-04-04
- Posts: 22

bob

Thanks so much. I get that perfectly. As an aside, can I calculate the speed that the heavier weight leaves the floor using conservation of momentum?

Offline

**bob bundy****Moderator**- Registered: 2010-06-20
- Posts: 7,016

hi jimi70,

In the idealised world where the strings are inelastic and there is no friction at the peg, yes you can.

Bob

Offline

**jimi70****Member**- Registered: 2012-04-04
- Posts: 22

Hi bob

Sorry to ressurect this, but can I use conservation of momentum with gravity acting on the sytem? I thought the condition was no external forces.

Offline

**bob bundy****Moderator**- Registered: 2010-06-20
- Posts: 7,016

hi jimi70,

Interesting question that got me thinking. I have two text books that cover this kind of work.

In 'Advanced Level Applied maths' by Lambe I found a very similar example where one mass (A) hangs over the edge of a table and the other (B) rests on the table. The string is initially slack, so only A moves.

When the string becomes taut, Lambe uses conservation of momentum to work out the new common velocity.

The second example is exactly like your problem with just different masses. It comes from 'A Shorter Intermediate Mechanics' by Humphrey and Topping. In the final stage of the motion they use conservation of momentum to work out the new velocity of the masses.

Bob

Offline

Pages: **1**