# Homework Help: Pulley Question, mechanics question

1. Oct 12, 2016

### Kajan thana

1. The problem statement, all variables and given/known data
Two particles A and B of mass 4kg and 3kg respectively are connected by a light inextensible string which passes over a small smooth pulley. The particles are released from rest with taut.

When A has travelled a distance of 2m it strikes the ground and immediately comes to rest.
Question: Assuming the B does hit the pulley find the greatest height that B reaches above its initial position.

2. Relevant equations
Kinematic equations

3. The attempt at a solution

I found out the the time it takes for the particle A to go down by 2m using this equation s=ut+0.5at^2.
The time is 1.69 rounded to two decimal place so if Particle A go down by 1.69 second thant the Particle B will go up by 1.69 second.

Now I know that for Particle B, U=0 A=-1.4 T=1.69 so using kinematics I can find the displace which will be -1.183m so the distance it goes by 1.183m. Which is a incorrect answer.

2. Oct 12, 2016

### Staff: Mentor

Why is the distance that particle B goes up different from the distance that particle A goes down in the 1.69 sec? Let's see your calculation of the acceleration.

3. Oct 12, 2016

### Staff: Mentor

Particle B is rising and accelerating upwards so long as the string is taut (particle A hasn't hit the ground). So when A hits the ground and the string goes slack, particle B will still have whatever velocity it achieved until that moment. It becomes a projectile launched with that initial speed upwards from that point. So when you wrote:

You were not considering the new projectile status of particle B. At the instant particle B becomes a free projectile, the motion from its travels while moving due to A's influence are behind it. You need to consider the new situation for B from that point on.

Edit: Ah, Chestermiller got there before me with a much more subtle approach

4. Oct 12, 2016

### Kajan thana

4g-t=4a and t-3g=3a solving it with elimination will get you a=g/7 same as 9.8/7

5. Oct 12, 2016

### Staff: Mentor

Good. So, since their accelerations are the same and their initial velocities are the same, how come they travel different distances in the same amount of time?

6. Oct 12, 2016

### Kajan thana

Thanks Mentor It made so much sense now after you clear explanation so just to make sure I understood the concept.
The acceleration I found is constant until it touches the ground after that it will change.

7. Oct 12, 2016

### Staff: Mentor

So, are you OK now, and have you achieved the answer you desire?

8. Oct 12, 2016

### Kajan thana

Yes I think I did mentor.

So there will be 2m distance caused by the particle A. Then when the particle A touches the ground the string no longer will taut so the acceleration can not be applied so I will apply a=-9.8 with the initial velocity as same as the instantaneous final velocity of particle A. and the final velocity will be 0. Using kinematics the answer will be 16/7.

Thanks again