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M1 Connected Particles Q

2

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

Reply 1

Eau

15

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...

Reply 2

Forté

2

what was the answer meant to be?

Reply 3

Tut

OP

2

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

Reply 4

Eau

15

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 :smile:

Reply 5

Tut

OP

2

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

Reply 6

Tut

OP

2

The answer is 0.528s !!! Tricky one eh!

Reply 7

Eau

15

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:

Reply 8

Tut

OP

2

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 :smile:

Reply 9

Forté

2

my only question is, was it from an AQA exam paper...
(please say no)

Reply 10

Codefusion

1

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)

Reply 11

Tut

OP

2

Lol.. Heinmann M1 book page 113 question 11...

Reply 12

Tut

OP

2

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!!!

Reply 13

Forté

2

can you explain why accelleration is 0? Earlier in the question it was worked out to be 5.71 :confused:

Reply 14

Mohit_C

10

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?

Reply 15

Codefusion

1

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.