Tut
Badges: 2
#1
Report Thread starter 13 years ago
#1
Hey guys, i can do the first two aprts of the question jsut the last 1 confuses me. If somebody could help i wud very much appreciate it:

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 wen the string breaks

0.715m

(b) the velocity of the particles when the string breaks

2.86ms

(c) the further time that elapses before A hits the floor, given that the table is 0.9m high.

This is what confuses me. I thought i had to find the time it takes to reach the pulley and add that to the time taken to fall the 0.9m to the floor. Might have used the wrong formulas or sumthing but it keeps comnig out wrong! Anyhelp wud be very much appreciated once agen!

Thanx
Tut
0
reply
Eau
Badges: 15
#2
Report 13 years ago
#2
(Original post by Tut)
Hey guys, i can do the first two aprts of the question jsut the last 1 confuses me. If somebody could help i wud very much appreciate it:

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 wen the string breaks

0.715m

(b) the velocity of the particles when the string breaks

2.86ms

(c) the further time that elapses before A hits the floor, given that the table is 0.9m high.

This is what confuses me. I thought i had to find the time it takes to reach the pulley and add that to the time taken to fall the 0.9m to the floor. Might have used the wrong formulas or sumthing but it keeps comnig out wrong! Anyhelp wud be very much appreciated once agen!

Thanx
Tut
Wait a sec...

I got

(a) u=0, a = 9.8 (because smooth), s=?, t=0.5

so s = 1.225 m


(b) v = 4.9 m/s

Check this first...
0
reply
Forté
Badges: 2
Rep:
?
#3
Report 13 years ago
#3
what was the answer meant to be?
0
reply
Tut
Badges: 2
#4
Report Thread starter 13 years ago
#4
The two answers i've given are right. The aprticle is not falling under the acceleration of gravity. It's on a horizontal table so F=ma.

T = 0.5a
0.7g - T = 0.7a

Adding the two gives 0.7g = 1.2a

a= 5.72

s= ut +1/2at^2
= (0) + (1/2)(5.72)(0.25)
= 0.715
0
reply
Eau
Badges: 15
#5
Report 13 years ago
#5
HOLY CRAP... this is like the hardest mechanics question i did....

I got 0.123 s (3 s. f.) for part (c) - probably got it wrong...

Will post my working out soon
0
reply
Tut
Badges: 2
#6
Report Thread starter 13 years ago
#6
For part b

v^2 = u^2 + 2as

with u = 0, a = 5.72 and s = 0.715

v^2 = 0 + 2 x 5.72 x 0.715
= 8.1796

v = 2.86
0
reply
Tut
Badges: 2
#7
Report Thread starter 13 years ago
#7
The answer is 0.528s !!! Tricky one eh!
0
reply
Eau
Badges: 15
#8
Report 13 years ago
#8
(Original post by Tut)
The two answers i've given are right. The aprticle is not falling under the acceleration of gravity. It's on a horizontal table so F=ma.

T = 0.5a
0.7g - T = 0.7a

Adding the two gives 0.7g = 1.2a

a= 5.72

s= ut +1/2at^2
= (0) + (1/2)(5.72)(0.25)
= 0.715
huh? I thought the table is smooth - hence no friction la ma...

so T should equal 0, no??? :confused:
0
reply
Tut
Badges: 2
#9
Report Thread starter 13 years ago
#9
Frictional force would equal zero. There would still be tension in the string. If its a smooth table then it just basically means there is no frictional force, but the tension in the string wud remain regardless.

And the answers are the same as the back of the book
0
reply
Forté
Badges: 2
Rep:
?
#10
Report 13 years ago
#10
my only question is, was it from an AQA exam paper...
(please say no)
0
reply
Codefusion
Badges: 1
Rep:
?
#11
Report 13 years ago
#11
c) the further time that elapses before A hits the floor, given that the table is 0.9m high.
just calculate the time before it leaves the table and is subjected to the gravity.
0.715 - 1 = 0.285
since accelration is 0 and risistance is 0
t = d/s
t = 0.285/2.86 = 57/572
now
a = 9.8
s = 0.9
u = 0
t = ?
4.9t2 = 0.9
t = sqrt(9/49)
time = sqrt(9/49) + 57/572 = 0.528s (3 s.f)
0
reply
Tut
Badges: 2
#12
Report Thread starter 13 years ago
#12
Lol.. Heinmann M1 book page 113 question 11...
0
reply
Tut
Badges: 2
#13
Report Thread starter 13 years ago
#13
(Original post by Codefusion)
c) the further time that elapses before A hits the floor, given that the table is 0.9m high.
just calculate the time before it leaves the table and is subjected to the gravity.
0.715 - 1 = 0.285
since accelration is 0 and risistance is 0
t = d/s
t = 0.285/2.86 = 57/572
now
a = 9.8
s = 0.9
u = 0
t = ?
4.9t2 = 0.9
t = sqrt(9/49)
time = sqrt(9/49) + 57/572 = 0.528s (3 s.f)
Thank yooooouuuu!!!
0
reply
Forté
Badges: 2
Rep:
?
#14
Report 13 years ago
#14
can you explain why accelleration is 0? Earlier in the question it was worked out to be 5.71 :confused:
0
reply
Mohit_C
Badges: 8
Rep:
?
#15
Report 13 years ago
#15
By the looks of it, acceleration might have been 0 as it had been travelling at a constand speed until the end of the table?
0
reply
Codefusion
Badges: 1
Rep:
?
#16
Report 13 years ago
#16
(Original post by domisakin)
can you explain why accelleration is 0? Earlier in the question it was worked out to be 5.71 :confused:
the table is smooth so there is no resistance , the string broke so there is nothing giving accelration. therefore it has no accelration or resistance ,so it's modeled as a particle with uniform velocity.
0
reply
AmarPatel98
Badges: 10
Rep:
?
#17
Report 4 years ago
#17
Why is the velocity 0 when it starts its verticle decent? (when it passes over the pulley and falls downwards)
0
reply
TenOfThem
Badges: 16
Rep:
?
#18
Report 4 years ago
#18
(Original post by AmarPatel98)
Why is the velocity 0 when it starts its verticle decent? (when it passes over the pulley and falls downwards)
There are rules regarding the resurrection of old threads - please start a thread of your own
0
reply
X

Quick Reply

Attached files
Write a reply...
Reply
new posts
Back
to top

University open days

  • Regent's University London
    Postgraduate Open Evening Postgraduate
    Thu, 19 Sep '19
  • Durham University
    Pre-Application Open Days Undergraduate
    Fri, 20 Sep '19
  • Loughborough University
    Undergraduate Open Day Undergraduate
    Fri, 20 Sep '19

What's your favourite genre?

Rock (171)
24.39%
Pop (174)
24.82%
Jazz (28)
3.99%
Classical (41)
5.85%
Hip-Hop (127)
18.12%
Electronic (45)
6.42%
Indie (115)
16.41%

Watched Threads

View All
Latest
My Feed