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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

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

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

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...

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

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???

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 = ?

4.9t

t = sqrt(9/49)

time = sqrt(9/49) + 57/572 = 0.528s (3 s.f)

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 = ?

4.9t^{2} = 0.9

t = sqrt(9/49)

time = sqrt(9/49) + 57/572 = 0.528s (3 s.f)

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 = ?

4.9t

t = sqrt(9/49)

time = sqrt(9/49) + 57/572 = 0.528s (3 s.f)

Thank yooooouuuu!!!

domisakin

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

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.

Why is the velocity 0 when it starts its verticle decent? (when it passes over the pulley and falls downwards)

Original post by AmarPatel98

Why is the velocity 0 when it starts its verticle decent? (when it passes over the pulley and falls downwards)

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