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M1 Help (pulley)

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.

Reply 1

SunGod87

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 ...

Reply 2

shimmerinapple

you got a right which is the distance. but b and c are wrong.

Reply 3

mizfissy815

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

Reply 4

shimmerinapple

^^ Thanks..that is correct.