A block of wood A of mass 0.5 kg rests on a rough horizontal table and is attached to one end of a light inextensible string.
The string passes over a small smooth pulley P fixed at the edge of the table.
The other end of the string is attached to a ball B of mass 0.8 kg which hangs freely below the pulley, as shown in Figure 4.
Block A experiences a frictional force of 3.68 N. The system is released from rest with the string taut.
After release, B descends a distance of 0.4 m
in 0.5 s.
Modelling A and B as particles, calculate;
( Figure 4 shows a pulley on the edge of the table, A is the one on the table with a mass of 0.5 kg and B is a small ball with a weight of 0.8 kg, the pulley is labelled P )
the acceleration of B,
B:
- S = 0.4
- u = 0
- t = 0.5
- a = ?
- S = ut + 1 / 2 at^2
- 0.4 = 1 / 2 a x ( 0.5 )^2
- 0.4 = 0.125a
- a = 3.2 m s^-2
The tension in the string,
( B = 0.8 kg )
( a = 3.2 m s^-2 )
B:
- F = M x a ( Force = mass x acceleration )
- ( When mass of particle is in Kg, the term “ g “ is include in the force of the particle )
- 0.8 g - T = ( 0.8 ) x ( 3.2 ) ( Ball is moving downwards, in the direction of its weight )
- T = ( 0.8 ) x ( 3.2 ) - 0.8 g ( g = 9.8, moving in the direction of gravity )
- T = -5.28
- T = 5.28 N
State how in your calculations you have used the information that the string is inextensible.
- A and B have the same amount of acceleration
Two particles P and Q have mass 0.5 kg and m kg respectively, where m < 0.5.
The particles are connected by a light inextensible string which passes over a smooth, fixed pulley.
Initially P is 3.15 m above horizontal ground.
The particles are released from rest with the string taut and the hanging parts of the string vertical, as shown in Figure 4.
After P has been descending for 1.5 s, it strikes the ground.
Particle P reaches the ground before Q has reached the pulley.
( Figure shows a pulley where particle P with a weight of 0.5 kg is higher than particle Q with a weight of Q, P is 3.15 m from the ground )
Show that the acceleration of P as it descends is 2.8 m s^-2.
P:
- S = 3.15
- u = 0
- a = ?
- t = 1.5
- S = ut + 1 / 2 at^2
- 3.15 = 1 / 2 a x ( 1.5 )^2
- 3.15 = 1.125a
- a = 2.8 m s ^-2
Find the tension in the string as P descends.
( Weight of P = 0.5 kg )
( a = 2.8 m s^-2 )
P:
- 0.5 g - T = ( 0.5 ) x ( 2.8 ) ( P is moving downwards, towards the weight of the particle )
- T = ( 0.5 ) x ( 2.8 ) - 0.5 g
- T = - 3.5
- T = 3.5 N
Show that m = 5 / 18
( Q = m kg )
( a = 2.8 m s^-2 )
( T = 3.5 N )
( Q is moving in the direction of the tension )
Q:
- T - m g = ( m ) x ( 2.8 )
- 3.5 - m g = 2.8m
- 3.5 = 2.8 m + m x ( 9.8 )
- 3.5 = 12.6m
- m = 5 / 18
State how you have used the information that the string is inextensible.
- Acceleration of P and Q are equal
When P strikes the ground, P does not rebound and the string becomes slack.
Particle Q then moves freely under gravity, without reaching the pulley, until the string becomes taut again.
Find the time between the instant when P strikes the ground and the instant when the string
becomes taut again.
( a = 2.8 )
( t = 1.5, when P hits the ground )
P:
- v = ?
- u = 0
- a = 2.8
- t = 1.5
- v = u + at
- v = ( 2.8 ) x ( 1.5 )
- v = 4.2
Q:
- v = 4.2
- u = - 4.2 ( The speed at which Q was originally at )
- a = - 9.8 ( Moves against gravity )
- t = ?
- v = u + at
- 4.2 = - 4.2 + ( 9.8 ) x t
- 8.4 = 9.8t
- t = 0.857 s
Two particles A and B, of mass m and 2m respectively, are attached to the ends of a light
inextensible string.
The particle A lies on a rough horizontal table. The string passes over a small smooth pulley P fixed on the edge of the table. The particle B hangs freely below the pulley, as shown in Figure 3.
Particle A experiences a frictional force of 2 / 3 m g.
The particles are released from rest with the string taut.
Immediately after release, the magnitude of the
acceleration of A and B is 4 / 9 g.
By writing down separate equations of motion for A and B,
( Figure shows a pulley on the edge of the table, A having a mass of m and B having a mass of 2m, the pulley is labelled P )
find the tension in the string immediately after the particles begin to move,
( a = 4 / 9 g )
B:
- 2m g - T = ( 2m ) x ( 4 / 9 g )
- T = ( 2m ) ( 4 / 9 g ) - 2m g
- T = - 10 / 9 m g
- T = 10 / 9 m g
When B has fallen a distance h, it hits the ground and does not rebound.
Particle A is then a distance 1 / 3 h from P.
Find the speed of A as it reaches P.
( T = 10 / 9 mg )
( a = 4 / 9 g )
A:
( The speed of which B hits the ground, which would be the speed of which A is 1 / 3 h away from P )
- u = 0
- v = ?
- a = 4 / 9 g
- S = h ( Both a and b move a total distance of h )
- v^2 = u^2 + 2as
- v^2 = 2 x ( 4 / 9 g ) x ( h )
- v^2 = 8 / 9 gh
- v = 2 root 2 / 3 gh
- f = ma
- 0 - 2 / 3 mg = ma ( Frictional force of 2 / 3, when B hits the ground, there’s no forward force, so T = 0 )
- a = - 2 / 3 g ( When B hits the ground, original acceleration is lost )
- S = 1 / 3 h ( Distance from P )
- u = 2 root 2 / 3 gh
- v = ?
- a = - 2 / 3 g
- v^2 = u^2 + 2as
- v^2 = 8 / 9 gh - 2 x ( - 2 / 3 g ) x ( 1 / 3 h )
- v^2 = 8 / 9 gh - 4 / 9 gh
- v^2 = 4 / 9 gh
- v = 2 / 3 root gh
State how you have used the information that the string is light.
- Same tension on A and B
Two particles A and B have masses 5m and km respectively, where k < 5.
The particles are connected by a light inextensible string which passes over a smooth light fixed pulley.
The system is held at rest with the string taut, the hanging parts of the string vertical and with A and B at the same height above a horizontal plane, as shown in Figure 4.
The system is released from rest.
After release, A descends with acceleration 1 / 4 g.
( Figure shows a pulley with A & B being at being same height above the ground )
Show that the tension in the string as A descends is 15 / 4 mg .
A:
- 5 mg - T = ( 5 m ) ( 1 / 4 g )
- T = - 15 / 4 mg
- T = 15 / 4 mg
Find the value of k.
( Two particles A and B have masses 5m and km respectively, where k < 5.
The particles are connected by a light inextensible string which passes over a smooth light fixed pulley.
The system is held at rest with the string taut, the hanging parts of the string vertical and with A and B at the same height above a horizontal plane, as shown in Figure 4.
The system is released from rest.
After release, A descends with acceleration 1 / 4 g.
( Figure shows a pulley with A & B being at being same height above the ground ) )
( T = 15 / 4 g )
B:
- T - km g = ( km ) x ( 1 / 4 g )
- 15 / 4m g - km g = 1 / 4 km g
- 15 / 4m g = 5 / 4 km g
- k = 3
State how you have used the information that the pulley is smooth.
- The tension in the two parts of the string is the same
After descending for 1.2 s, the particle A reaches the plane.
It is immediately brought to rest by the impact with the plane.
The initial distance between B and the pulley is such that, in the subsequent motion, B does not reach the pulley.
Find the greatest height reached by B above the plane.
- ( k = 3 )
- ( a = 1 / 4 g )
A:
- s =?
- u = 0
- a = 1 / 4 g
- t = 1.2
- s = ut + 1 / 2 at^2
- s = 1 / 2 x ( 1 / 4 g ) x ( 1.2 )^2
- s = 0.18 g ( height of A above the ground after descending )
- v = ?
- u = 0
- a = 1 / 4 g
- t = 1.2
- v = u + at
- v = ( 1/4 g ) x ( 1.2 )
- v = 0.3 g ( final velocity of A = final velocity at B )
B:
- v = 0.3 g
- u = 0
- a = - g
- ( B is under the influence of gravity after A no longer moves )
- s = s
- v^2 = u^2 + 2as
- ( 0.3 g )^2 = 2gs
- s = ( 0.3 g )^2 / 2g
- s = 0.441
- s = 2 x ( 0.18 ) x ( 9.8 ) + 0.441 ( 2 x ( 0.18 g ) + 0.441 )
- ( 0.18 g is x 2 because particle B is above the ground and the length travelled by A )
- s = 3.969
- s = 4 m
Two particles A and B have mass 0.4 kg and 0.3 kg respectively. The particles are attached to the ends of a light inextensible string. The string passes over a small smooth pulley which is fixed above a horizontal floor.
Both particles are held, with the string taut, at a height of 1 m
above the floor, as shown in Figure 3.
The particles are released from rest and in the subsequent motion B does not reach the pulley.
( Figure shows particles A and B 1 m above the ground on a pulley )
Find the tension in the string immediately after the particles are released.
A:
- 0.4 g - T = ( 0.4 ) x ( a )
- ( 0.4 it’s heavier than 0.3 so it falls more towards particle A )
B:
- T - 0.3 g = ( 0.3 ) x ( a )
- ( Solve the simultaneous equation )
- ( Due to opposite T values, we add the equations )
- 0.1 g = 0.7a
- a = 1.4 ms^-2
- 0.4 g - T = ( 0.4 ) x ( 1.4 )
- T = - 3.36
- T = 3.36 N
Find the acceleration of A immediately after the particles are released.
( Two particles A and B have mass 0.4 kg and 0.3 kg respectively. The particles are attached to the ends of a light inextensible string. The string passes over a small smooth pulley which is fixed above a horizontal floor.
Both particles are held, with the string taut, at a height of 1 m
above the floor, as shown in Figure 3.
The particles are released from rest and in the subsequent motion B does not reach the pulley.
( Figure shows particles A and B 1 m above the ground on a pulley ) )
- a = 1.4 ms^-2
- 0.1 g = 0.7a
- a = 1.4 ms^-2
- ( I would say to show the simultaneous equation from the previous question )
- ( A:
- 0.4 g - T = ( 0.4 ) x ( a )
- ( 0.4 it’s heavier than 0.3 so it falls more towards particle A )
B:
- T - 0.3 g = ( 0.3 ) x ( a )
- ( Solve the simultaneous equation )
- ( Due to opposite T values, we add the equations )
- 0.1 g = 0.7a
- a = 1.4 ms^-2 )
When the particles have been moving for 0.5 s, the string breaks.
Find the further time that elapses until B hits the floor.
- ( Two particles A and B have mass 0.4 kg and 0.3 kg respectively. The particles are attached to the ends of a light inextensible string. The string passes over a small smooth pulley which is fixed above a horizontal floor.
Both particles are held, with the string taut, at a height of 1 m
above the floor, as shown in Figure 3.
The particles are released from rest and in the subsequent motion B does not reach the pulley.
( Figure shows particles A and B 1 m above the ground on a pulley ) )
- ( a = 1.4 ms^-2 )
B:
- v = ?
- u = 0
- a = 1.4
- t = 0.5
- v = u + at
- v = ( 1.4 ) x ( 0.5 )
- v = 0.7 ( final velocity = initial velocity at B’s greatest height )
- s = ?
- u = 0
- a = 1.4
- t = 0.5
- s = ut + 1 / 2at^2
- s = 1 / 2 ( 1.4 ) x ( 0.5 )^2
- s = 0.175 ( height reached before string breaks )
A:
- s = 1 + 0.175 = 1.175 ( A is 1 m above the ground )
B:
- s = - 1.175 ( total displacement of B, which is why it’s negative )
- u = 0.7
- t = ?
- a = - 9.8
- s = ut + 1 / 2at^2
- 1.175 = 0.7t + 1/2 ( - 9.8 ) t^2
- 4.9t^2 - 0.7t - 1.175 = 0
- a = 4.9
- b = - 0.7
- c = - 1.175
- ( use of quadratic formula )
- t = 0.5663002307
- t = - 0.423430879
- ( t cannot be negative )
- t = 0.57
-
Proof by contradiction ( Algebratic Methods )12
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Trigonometric Equations and Identities7
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Modelling with differentiation2
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Pulleys18
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Binomial expansion12
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Proof By Exhaustion10
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Simplifying Rational Expressions6
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Sketching Graphs2
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The Discriminant3
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The Factor Theorem3
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Partial Fractions5
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Representations Of Data32
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Sampling Methods24
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Hypothesis Testing18
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Statistical Distribution12
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Sequences And Series33
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Moments14
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Functions And Graphs4
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Trigonometry7
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Radians2
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Forces And Friction4
