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I am revising past papers for my mechanics exam and I am stuck on the last question.

Two particles P and Q, of mass 4kg and 6kg respectively, are joined by a light inextensible string. Initially the particles are at rest on a rough horizontal plane with the string taut. The coefficient of friction between each particle and the plane is 2/7. A constant force of magnitude 40 N is then applied to Q in the direction PQ.

(a) Show that the acceleration of Q is 1.2ms-².

(b) Calculate the tenson in the string when the system is moving.

After the particles have been moving for 7s, the string breaks. The particle Q remains under the actionof the force of magnitude 40N.

(d) Show that P continues to move for a further 3 seconds.

(e) Calculate he speed of Q at the instant when P com to rest.

I know I know how to these questions, I might be being silly and doing careless mistakes.

Any help is useful, Thanks.

Two particles P and Q, of mass 4kg and 6kg respectively, are joined by a light inextensible string. Initially the particles are at rest on a rough horizontal plane with the string taut. The coefficient of friction between each particle and the plane is 2/7. A constant force of magnitude 40 N is then applied to Q in the direction PQ.

(a) Show that the acceleration of Q is 1.2ms-².

(b) Calculate the tenson in the string when the system is moving.

After the particles have been moving for 7s, the string breaks. The particle Q remains under the actionof the force of magnitude 40N.

(d) Show that P continues to move for a further 3 seconds.

(e) Calculate he speed of Q at the instant when P com to rest.

I know I know how to these questions, I might be being silly and doing careless mistakes.

Any help is useful, Thanks.

MightyMeanie

I am revising past papers for my mechanics exam and I am stuck on the last question.

Two particles P and Q, of mass 4kg and 6kg respectively, are joined by a light inextensible string. Initially the particles are at rest on a rough horizontal plane with the string taut. The coefficient of friction between each particle and the plane is 2/7. A constant force of magnitude 40 N is then applied to Q in the direction PQ.

(a) Show that the acceleration of Q is 1.2ms-².

(b) Calculate the tenson in the string when the system is moving.

After the particles have been moving for 7s, the string breaks. The particle Q remains under the actionof the force of magnitude 40N.

(d) Show that P continues to move for a further 3 seconds.

(e) Calculate he speed of Q at the instant when P com to rest.

I know I know how to these questions, I might be being silly and doing careless mistakes.

Any help is useful, Thanks.

Two particles P and Q, of mass 4kg and 6kg respectively, are joined by a light inextensible string. Initially the particles are at rest on a rough horizontal plane with the string taut. The coefficient of friction between each particle and the plane is 2/7. A constant force of magnitude 40 N is then applied to Q in the direction PQ.

(a) Show that the acceleration of Q is 1.2ms-².

(b) Calculate the tenson in the string when the system is moving.

After the particles have been moving for 7s, the string breaks. The particle Q remains under the actionof the force of magnitude 40N.

(d) Show that P continues to move for a further 3 seconds.

(e) Calculate he speed of Q at the instant when P com to rest.

I know I know how to these questions, I might be being silly and doing careless mistakes.

Any help is useful, Thanks.

a) first treat as 1 paritcle as tention in string will cancel.

draw a good diagram

apply newtons 2nd law (f=ma) horizontally and end up with an equation with 2 unknowns the frictional force (F) and acceleration

to find friction resolve vertically (f=ma) and then using the value u get for R use F=(mu)R and mu as 2/7

put new value for F into first equation.

b) do same as before but just use particle p or q ill use p.

draw diagram of forces acting on P

get equation ie T-F=ma

sub in all values

c) find speed of P at seven seconds use suvat assume constant acceleration and start at 0ms^-1

c)......... then draw another diagram for P but not with the T pulling it farward then show decelleration using the same method as before (F=ma) then apply the new deceleration to suvat with

a=thing u just got

u=value of speed at 7s

v=0

t=t

v=u+at

d)again much the same use speed of P at 7s as will be same as Q, then again draw diagram of Q but without T will probably accelerate do F=ma again and then suvat the acceleration as u know initial speed, acceleration and time

and that is that

a=thing u just got

u=value of speed at 7s

v=0

t=t

v=u+at

d)again much the same use speed of P at 7s as will be same as Q, then again draw diagram of Q but without T will probably accelerate do F=ma again and then suvat the acceleration as u know initial speed, acceleration and time

and that is that

amo1

c)......... then draw another diagram for P but not with the T pulling it farward then show decelleration using the same method as before (F=ma) then apply the new deceleration to suvat with

a=thing u just got

u=value of speed at 7s

v=0

t=t

v=u+at

d)again much the same use speed of P at 7s as will be same as Q, then again draw diagram of Q but without T will probably accelerate do F=ma again and then suvat the acceleration as u know initial speed, acceleration and time

and that is that

a=thing u just got

u=value of speed at 7s

v=0

t=t

v=u+at

d)again much the same use speed of P at 7s as will be same as Q, then again draw diagram of Q but without T will probably accelerate do F=ma again and then suvat the acceleration as u know initial speed, acceleration and time

and that is that

THANK YOU SO MUCH

amo1

a) first treat as 1 paritcle as tention in string will cancel.

draw a good diagram

apply newtons 2nd law (f=ma) horizontally and end up with an equation with 2 unknowns the frictional force (F) and acceleration

to find friction resolve vertically (f=ma) and then using the value u get for R use F=(mu)R and mu as 2/7

put new value for F into first equation.

b) do same as before but just use particle p or q ill use p.

draw diagram of forces acting on P

get equation ie T-F=ma

sub in all values

c) find speed of P at seven seconds use suvat assume constant acceleration and start at 0ms^-1

draw a good diagram

apply newtons 2nd law (f=ma) horizontally and end up with an equation with 2 unknowns the frictional force (F) and acceleration

to find friction resolve vertically (f=ma) and then using the value u get for R use F=(mu)R and mu as 2/7

put new value for F into first equation.

b) do same as before but just use particle p or q ill use p.

draw diagram of forces acting on P

get equation ie T-F=ma

sub in all values

c) find speed of P at seven seconds use suvat assume constant acceleration and start at 0ms^-1

I did that originally and got the wrong answer, and did it that way again and still got the wrong answer. After thinking about it and playing around with numbers i finally got it. You dont treat the tension as cancelling.

Here is what i did to get the right answer.

40 - (friction for Q) - T = 6a → Q

T - (friction for P) = 4a → P

which gives

40 - 16.8 - T = 6a

T - 11.2 = 4a

add the two together which cancels out the T

12 = 10a

a = 1.2

T - 11.2 = 4a

T = 4 x 1.2 + 11.2

T= 16

i am still stuck on d and e though, I cant get my head around them... i will keep trying.

d) after 7 secs speed of the two particles is 7*1.2 = 8.4 m/s

when string breaks there is only friction acting against P so it decellerates.

friction force = 4*9.8*2/7 = 11.2 N

deceleration d = 11.2/4 = 2.8 m/s² (Fr=md)

v=u+at

0 = 8.4 - 2.8t

t = 8.4/2.8

t = 3 s

=====

e)

Friction against Q is 6*9.9*2/7 = 16.8 N

T-Fr = ma

40 - 16.8 = 6a

a = 23.2/6

a = 3.867 m/s²

v = u + at

v = 8.4 + 3.867*3

v = 8.4 + 11.6

v = 20 m/s

========

when string breaks there is only friction acting against P so it decellerates.

friction force = 4*9.8*2/7 = 11.2 N

deceleration d = 11.2/4 = 2.8 m/s² (Fr=md)

v=u+at

0 = 8.4 - 2.8t

t = 8.4/2.8

t = 3 s

=====

e)

Friction against Q is 6*9.9*2/7 = 16.8 N

T-Fr = ma

40 - 16.8 = 6a

a = 23.2/6

a = 3.867 m/s²

v = u + at

v = 8.4 + 3.867*3

v = 8.4 + 11.6

v = 20 m/s

========

Fermat

d) after 7 secs speed of the two particles is 7*1.2 = 8.4 m/s

when string breaks there is only friction acting against P so it decellerates.

friction force = 4*9.8*2/7 = 11.2 N

deceleration d = 11.2/4 = 2.8 m/s² (Fr=md)

v=u+at

0 = 8.4 - 2.8t

t = 8.4/2.8

t = 3 s

=====

e)

Friction against Q is 6*9.9*2/7 = 16.8 N

T-Fr = ma

40 - 16.8 = 6a

a = 23.2/6

a = 3.867 m/s²

v = u + at

v = 8.4 + 3.867*3

v = 8.4 + 11.6

v = 20 m/s

========

when string breaks there is only friction acting against P so it decellerates.

friction force = 4*9.8*2/7 = 11.2 N

deceleration d = 11.2/4 = 2.8 m/s² (Fr=md)

v=u+at

0 = 8.4 - 2.8t

t = 8.4/2.8

t = 3 s

=====

e)

Friction against Q is 6*9.9*2/7 = 16.8 N

T-Fr = ma

40 - 16.8 = 6a

a = 23.2/6

a = 3.867 m/s²

v = u + at

v = 8.4 + 3.867*3

v = 8.4 + 11.6

v = 20 m/s

========

Thank you

Thanks with your help so far, I now know how to do simular questions like the back of my hand.

I consider that I am ok with momentum, I have got all of my questions correct. However, this question I keep on getting the wrong answer.

A Particle P of mass 2kg is moving with speed u ms-¹ in a straight line on a smooth horizontal plane. The particle P collides directly with a particle Q of mass 4kg which is at rest on the same horizontal plane. Imediately after the collision P and Q are moving in opposite directions and the speed of P is one-third the speed of Q.

Show that the speed of P immediately after the collision is 1/5u ms-¹.

However when I did this I worked it as 3/5u ms-¹. I have left it a day and repeated, but i am still getting 3/5u ms-¹.

Here is how I am doing it.

BEFORE AFTER

Momentum P + Momentum Q = Momentum P + Momentum Q

(2 x u ) + ( 4 x 0 ) = (2 x - ⅓V) + (4 x V)

2u = 3⅓ V

V = 3/5u ms-¹

I consider that I am ok with momentum, I have got all of my questions correct. However, this question I keep on getting the wrong answer.

A Particle P of mass 2kg is moving with speed u ms-¹ in a straight line on a smooth horizontal plane. The particle P collides directly with a particle Q of mass 4kg which is at rest on the same horizontal plane. Imediately after the collision P and Q are moving in opposite directions and the speed of P is one-third the speed of Q.

Show that the speed of P immediately after the collision is 1/5u ms-¹.

However when I did this I worked it as 3/5u ms-¹. I have left it a day and repeated, but i am still getting 3/5u ms-¹.

Here is how I am doing it.

BEFORE AFTER

Momentum P + Momentum Q = Momentum P + Momentum Q

(2 x u ) + ( 4 x 0 ) = (2 x - ⅓V) + (4 x V)

2u = 3⅓ V

V = 3/5u ms-¹

SunGod

You're not setting one direction to be positive and the other direction to be negative, I remember getting stuck in the same place.

sorry, that was a typo.. i had used positve and negative

Principle of Conservation Of Linear Momentum:

mpup+mquq=mpvp+mqvq

Remember the direction of p is reversed after the collision. Since we are dealing with velocities, which are vectors, the velocity of p therefore becomes negative. Hence:

2u=(-2(1/3))v+4v

Multiply out by 3:

6u=-2v+12v

=>6u=10v

=>v=((6/10)um(s^-1))=((3/5)um(s^-1))

But that is the velocity of q after the collision. It says that the velocity of p is ((1/3)vq). Therefore:

vp=((1/3)vq)=(((1/3)(3/5))u)=((3/15)v)=((1/5)v) Q. E. D.

Newton.

mpup+mquq=mpvp+mqvq

Remember the direction of p is reversed after the collision. Since we are dealing with velocities, which are vectors, the velocity of p therefore becomes negative. Hence:

2u=(-2(1/3))v+4v

Multiply out by 3:

6u=-2v+12v

=>6u=10v

=>v=((6/10)um(s^-1))=((3/5)um(s^-1))

But that is the velocity of q after the collision. It says that the velocity of p is ((1/3)vq). Therefore:

vp=((1/3)vq)=(((1/3)(3/5))u)=((3/15)v)=((1/5)v) Q. E. D.

Newton.

Newton

Principle of Conservation Of Linear Momentum:

mpup+mquq=mpvp+mqvq

Remember the direction of p is reversed after the collision. Since we are dealing with velocities, which are vectors, the velocity of p therefore becomes negative. Hence:

2u=(-2(1/3))v+4v

Multiply out by 3:

6u=-2v+12v

=>6u=10v

=>v=((6/10)um(s^-1))=((3/5)um(s^-1))

But that is the velocity of q after the collision. It says that the velocity of p is ((1/3)v). Therefore:

vp=((1/3)v)=(((1/3)(3/5))u)=((3/15)v)=((1/5)v) Q. E. D.

Newton.

mpup+mquq=mpvp+mqvq

Remember the direction of p is reversed after the collision. Since we are dealing with velocities, which are vectors, the velocity of p therefore becomes negative. Hence:

2u=(-2(1/3))v+4v

Multiply out by 3:

6u=-2v+12v

=>6u=10v

=>v=((6/10)um(s^-1))=((3/5)um(s^-1))

But that is the velocity of q after the collision. It says that the velocity of p is ((1/3)v). Therefore:

vp=((1/3)v)=(((1/3)(3/5))u)=((3/15)v)=((1/5)v) Q. E. D.

Newton.

ahhhhhhhhhhhhhhhhhhhhhhhhhhhh, I see. Thank you very much

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