# M1 Help (pulley)Watch

Thread starter 12 years ago
#1
Need help with M1 question excercise 5D question 11.

Two particles A and B of masses 0.5kg and 0.7kg respectively are connected by a light inextensible string. Particle A on a smooth horizontal table 1m from a fixed smooth pulley at the edge of the table. The string passes over the pulley and B hangs freely 0.75m above the floor. The system is released from rest with the string taut. After 0.5s the string breaks. Find a) the distance the prticles have moved when the string breaks. b) the velocity of the particles have moved when the string breaks. c)the further time that elapses before A reaches the floor, given that the table is 0.9m high.

Thanks.
0
12 years ago
#2
Equation of motion of particle B:
F = ma
F = 0.7*9.8
F = 6.86 N

This is the force acting on particle A:
F = ma
6.86 = 0.5*a
a = 13.72 m/s^2

Using kinematic equations:
s = ut + 1/2at^2
s = 0*0.5 + (0.5)(13.72)(0.5^2)
s = (0.5)(13.72)(0.25)
s = 6.86*0.25
s = 1.715 m

v^2 = u^2 + 2as
v^2 = 0^2 + 2(13.72)(1.715)
v^2 = 47.0596
v = 6.86 m/s

At 0.5 s particle B is 1.715m from the table, thus to get to the floor it needs to travel (9 - 1.715)m = 7.285m

We also know the speed it is travelling at at this given time (worked out v) and the fact that it is accelerating due to gravity alone.

s = ut + 1/2at^2
7.285 = 6.86t + (0.5)(9.8)t^2
7.285 = 6.86t + 4.9t^2
4.9t^2 + 6.86t - 7.285 = 0

Solving as a quadratic for t:

[-b ± SQRT(b^2 - 4ac)]/2a
-6.86 ± SQRT((6.86)^2 - (4)(4.9)(-7.285))]/(2)(4.9)

t = 0.705963973
t = -2.105963973

We neglect the negative t value since we want to know the "further" time at which this happens. Thus t = 0.706 s

Edit: Please tell me this is correct ...
0
Thread starter 12 years ago
#3
you got a right which is the distance. but b and c are wrong.
0
12 years ago
#4
a) Before string breaks:
Equation of motion for A:
F = ma
T = 0.5a .....eq. 1

Equation of motion for B:
0.7g – T = 0.7a....eq.2
Add 1 and 2:
0.7g = 1.2a
a = (0.7 x 9.8) / 1.2 = 5.716
u = 0 ms–1   a = 5.716 ms–2  t = 0.5 s
s = ut + ½ at2
s= 0 + ½ × 5.716 × 0.5 2
s = 0.715 m

b) v = u + at
v = 0 + 5.716 × 0.5 = 2.858
The particles are moving at 2.86 ms–1 when the string breaks

c) When string breaks A is (1 – 0.715) m = 0.285 m from the edge of the table.
So time to edge = distance/ speed = (0.285/ 2.858) = 0.09972 s

Moving as projectile:
Vertical motion:
u = 0 ms–1   s =  0.9 m a = 9.8 ms–2
s = ut + ½ at2
0.9 = 0 + ½ × 9.8 t2
t2= 0.9/ 4.9
t = ± 0.4285…..t = 0.4895
Total time = (0.099 72 + 0.4285)
s = 0.528 s
0
Thread starter 12 years ago
#5
^^ Thanks..that is correct.
0
X

new posts

Latest
My Feed

### Oops, nobody has postedin the last few hours.

Why not re-start the conversation?

see more

### See more of what you like onThe Student Room

You can personalise what you see on TSR. Tell us a little about yourself to get started.

### University open days

• University of East Anglia
All Departments Open 13:00-17:00. Find out more about our diverse range of subject areas and career progression in the Arts & Humanities, Social Sciences, Medicine & Health Sciences, and the Sciences. Postgraduate
Wed, 30 Jan '19
• Aston University
Wed, 30 Jan '19
• Solent University
Careers in maritime Undergraduate
Sat, 2 Feb '19

### Poll

Join the discussion

Remain (863)
80.28%
Leave (212)
19.72%