M1 pulleys question.

Two questions I just can't get my head round.

11) Two particles A and B of masses 0.5kg and 0.7kg respectively are connected by a light inextensible string. Particle A rests 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 particles have moved when the string breaks.
(b) the velocity of the particles when the string breaks.
(c) the further time that elapses before A reaches the floor, given that the table is 0.9m high.

12) A particle P of mass 3kg lies on a smooth horizontal table top which is 1.5m above the floor. A light inextensible string of length 1m connects P to a particle Q, also of mass 3kg, which hangs freely over a small smooth pulley at the edge of the table. Initially P is held at rest at a point 0.5m from the pulley. When the system is released from rest find;
a) the speed of P when it reaches the pulley.
b) the tension in the string.

Answers.
11a - 0.715m
11b - 2.86ms^-2
11c - 0.528s

12a - 2.21ms^-1
12b - 14.7N

:confused:

Reply 1

The Chameleon

OP

Anyone? Please.

Reply 2

hihihihi

The Chameleon

Two questions I just can't get my head round.

11) Two particles A and B of masses 0.5kg and 0.7kg respectively are connected by a light inextensible string. Particle A rests 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 particles have moved when the string breaks.
(b) the velocity of the particles when the string breaks.
(c) the further time that elapses before A reaches the floor, given that the table is 0.9m high.

Answers.
11a - 0.715m
11b - 2.86ms^-2
11c - 0.528s

for A: T=0.5a ......................(1)
for B: 0.7g-T=0.7a..................(2)

(1)+(2)

0.7g=1.2a
a=1.7g / 1.2
a=5.72ms^-2

after 0.5 seconds

s=ut+ 0.5at^2
s= 0.5* 5.72*0.5^2
s=0.715 m

Reply 3

C4>O7

5

[][]

Mechanics 2.doc21.5KB

Reply 4

hihihihi

The Chameleon

Two questions I just can't get my head round.

11) Two particles A and B of masses 0.5kg and 0.7kg respectively are connected by a light inextensible string. Particle A rests 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 particles have moved when the string breaks.
(b) the velocity of the particles when the string breaks.
(c) the further time that elapses before A reaches the floor, given that the table is 0.9m high.

Answers.
11a - 0.715m
11b - 2.86ms^-2
11c - 0.528s

b) very simple after part a) :
uvsat it

v=u+at
v=0 + 5.72*0.5
v=2.86ms^-1

c)Once the string breaks, there is no tension. But because the table is smooth, A will continue to travel at 2.86ms^-1 until it reaches the end of the table.

so the time it takes for A to reach end of table:
s = 1-0.715
s=0.285m

s=vt
t=s/v
t=0.285/2.86
t=0.0997s

then time it takes for A to reach floor:
u=0
s=0.9m
a=g
t=t

s=0.5*at^2
t= sqrt [0.9/(0.5*9.8)]
t=0.429s

Total time =0.429s + 0.0997s
=0.529s

[attachment for part a)]

pulley.JPG8.6KB

Reply 5

The Chameleon

OP

C4>O7

[][]

Thankyou, but the answer for t in the book is 0.528s

Reply 6

username9816

14

The Chameleon

11) Two particles A and B of masses 0.5kg and 0.7kg are connected by a light inextensible string. Particle A rests 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 particles have moved when the string breaks.
(b) the velocity of the particles when the string breaks.
(c) the further time that elapses before A reaches the floor, given that the table is 0.9m high.

12) A particle P of mass 3kg lies on a smooth horizontal table top which is 1.5m above the floor. A light inextensible string of length 1m connects P to a particle Q, also of mass 3kg, which hangs freely over a small smooth pulley at the edge of the table. Initially P is held at rest at a point 0.5m from the pulley. When the system is released from rest find;
a) the speed of P when it reaches the pulley.
b) the tension in the string.

11.) a.) For A: F = ma ---> T = 0.5a
For B: F = ma ---> 0.7g - T = 0.7a ---> T = 0.7g - 0.7a

Equating the expressions for T gives:
---> 0.5a = 0.7g - 0.7a
---> 1.2a = 0.7g
---> a = [(0.7)(9.8)]/1.2
---> a = 5.717 ms^-2.

Until String Breaks:
a = 5.717, u = 0, t = 0.5 s.

s = ut + 1/2at^2
---> s = 1/2(5.717)(0.5)^2
---> Distance Moved = 0.7146 m

b.) At instant of string braking:
v = u + at
---> v = 5.717 * 0.5
---> Velocity Of Particles = 2.86 ms^-1

c.) From point of string braking, until A reaches the edge of the table:
s = 1 - 0.7146 = 0.2854 m, a = 0, v and u = 2.86.

s = ut + 1/2at^2
---> 0.2854 = 2.86t
---> t = 0.1 s

For A reaching the floor from the edge of the table:
a = 9.8, u = 0, s = 0.9.

s = ut + 1/2at^2
---> 0.9 = 4.9t^2
---> t = Sq.rt (0.9/4.9)
---> t = 0.4286 s

Time Elapsed (From Point Of String Braking) = 0.1 + 0.4286 = 0.5286 s

12.) For P: F = ma ---> T = 3a
For Q: F = ma ---> 3g - T = 3a ---> T = 3g - 3a

Equating the expressions for T gives:
---> 3a = 3g - 3a
---> 6a = 3g
---> 2a = g
---> a = g/2 ms^-2.

When the system is released from rest, for P reaching the pulley:
a = g/2, s = 0.5 m, u = 0.

v^2 = u^2 + 2as
---> v^2 = 2(g/2)(0.5)
---> v^2 = g/2
---> v = Sq.rt (g/2)
---> Speed Of P (Reach Pulley) = 2.21 ms^-1

b.) T = 3g - 3a
---> T = 29.4 - 3(g/2)
---> Tension (String) = 14.7 N

Reply 7

hihihihi

Nima

For A reaching the floor from the edge of the table:
a = 9.8, u = 2.86, s = 0.9.

s = ut + 1/2at^2
---> 0.9 = 2.86t + 4.9t^2
---> 4.9t^2 + 2.86t - 0.9 = 0
---> t = {-2.96 + Sq.rt[2.86^2 - 4(4.9)(-0.9)]}/9.8
---> t = 0.216 s

Time Elapsed (From Point Of String Braking) = 0.1 + 0.216 = 0.316 s

vertical u=0ms^-1

Reply 8

hihihihi

Nima

it'll start the drop at 2.86 ms^-1 though, it won't drop from rest. I'm re-doing part c anyway.