4'1a Further Mechanics Momentum concepts

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4.1a Further MechanicsMomentum concepts

• Breithaupt pages 4 to 17

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AQA A2 Specification
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Momentum, p

• momentum mass x velocity
• p mv
• m in kilograms (kg)
• v in metres per second (ms-1)
• p in kilograms metres per second (kg ms-1)
• Momentum is a VECTOR quantity
• direction the same as the velocity

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Momentum, p

• momentum mass x velocity
• p mv
• m in kilograms (kg)
• v in metres per second (ms-1)
• p in kilograms metres per second (kg ms-1)
• Momentum is a VECTOR quantity
• direction the same as the velocity

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Momentum, p

• momentum mass x velocity
• p mv
• m in kilograms (kg)
• v in metres per second (ms-1)
• p in kilograms metres per second (kg ms-1)
• Momentum is a VECTOR quantity
• direction the same as the velocity

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Momentum, p

• momentum mass x velocity
• p mv
• m in kilograms (kg)
• v in metres per second (ms-1)
• p in kilograms metres per second (kg ms-1)
• Momentum is a VECTOR quantity
• direction the same as the velocity

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Newtons 2nd law (A2 version)

• The resultant force acting on an object is
  proportional to the rate of change of momentum of
  the object and is in the same direction as the
  resultant force.
• SF a ?(p)
• ?(t)

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Newtons 2nd law (A2 version)

• The resultant force acting on an object is
  proportional to the rate of change of momentum of
  the object and is in the same direction as the
  resultant force.
• SF a ?(p)
• ?(t)

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• Inserting a constant of proportionality k
• SF k ?(p)
• ?(t)
• but p mv
• hence SF k ?(mv)
• ?(t)
• If the mass, m remains constant
• SF k m ?(v)
• ?(t)

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• Inserting a constant of proportionality k
• SF k ?(p)
• ?(t)
• but p mv
• hence SF k ?(mv)
• ?(t)
• If the mass, m remains constant
• SF k m ?(v)
• ?(t)

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• Inserting a constant of proportionality k
• SF k ?(p)
• ?(t)
• but p mv
• hence SF k ?(mv)
• ?(t)
• If the mass, m remains constant
• SF k m ?(v)
• ?(t)

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• but ?(v) a (acceleration)
• ?(t)
• hence SF k m a
• A force of one newton is defined as that required
  to cause an acceleration of 1 ms -2 with a mass
  of 1 kg.
• Inserting these values into SF k m a
• gives 1 k x 1 x 1
• and so k 1
• giving SF m a (the AS version of Newtons
  2nd law)
• Note This simplified version only applies for an
  object of constant mass.

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• but ?(v) a (acceleration)
• ?(t)
• hence SF k m a
• A force of one newton is defined as that required
  to cause an acceleration of 1 ms -2 with a mass
  of 1 kg.
• Inserting these values into SF k m a
• gives 1 k x 1 x 1
• and so k 1
• giving SF m a (the AS version of Newtons
  2nd law)
• Note This simplified version only applies for an
  object of constant mass.

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• but ?(v) a (acceleration)
• ?(t)
• hence SF k m a
• A force of one newton is defined as that required
  to cause an acceleration of 1 ms -2 with a mass
  of 1 kg.
• Inserting these values into SF k m a
• gives 1 k x 1 x 1
• and so k 1
• giving SF m a (the AS version of Newtons
  2nd law)
• Note This simplified version only applies for an
  object of constant mass.

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• but ?(v) a (acceleration)
• ?(t)
• hence SF k m a
• A force of one newton is defined as that required
  to cause an acceleration of 1 ms -2 with a mass
  of 1 kg.
• Inserting these values into SF k m a
• gives 1 k x 1 x 1
• and so k 1
• giving SF m a (the AS version of Newtons
  2nd law)
• Note This simplified version only applies for an
  object of constant mass.

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• but ?(v) a (acceleration)
• ?(t)
• hence SF k m a
• A force of one newton is defined as that required
  to cause an acceleration of 1 ms -2 with a mass
  of 1 kg.
• Inserting these values into SF k m a
• gives 1 k x 1 x 1
• and so k 1
• giving SF m a (the AS version of Newtons
  2nd law)
• Note This simplified version only applies for an
  object of constant mass.

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• but ?(v) a (acceleration)
• ?(t)
• hence SF k m a
• A force of one newton is defined as that required
  to cause an acceleration of 1 ms -2 with a mass
  of 1 kg.
• Inserting these values into SF k m a
• gives 1 k x 1 x 1
• and so k 1
• giving SF m a (the AS version of Newtons
  2nd law)
• Note This simplified version only applies for an
  object of constant mass.

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Force and Momentum

• Force is equal to the rate of change of momentum.
• F ?(mv) / ?t
• F in newtons (N)
• ?(mv) in kilograms metres per second (kg ms-1)
• ?t in seconds (s)

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Force and Momentum

• Force is equal to the rate of change of momentum.
• F ?(mv) / ?t
• F in newtons (N)
• ?(mv) in kilograms metres per second (kg ms-1)
• ?t in seconds (s)

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Force and Momentum

• Force is equal to the rate of change of momentum.
• F ?(mv) / ?t
• F in newtons (N)
• ?(mv) in kilograms metres per second (kg ms-1)
• ?t in seconds (s)

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Question 1

• A car of mass 800 kg moving at a velocity of 30
  ms -1 is brought to rest by a braking force of
  1200 N.
• Calculate
• (a) its initial momentum
• (b) the time taken to stop the car.

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Question 1

• A car of mass 800 kg moving at a velocity of 30
  ms -1 is brought to rest by a braking force of
  1200 N.
• Calculate
• (a) its initial momentum
• (b) the time taken to stop the car.

• (a) p mv
• 800 kg x 30 ms-1
• momentum 24 000 kg ms-1

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Question 1

• A car of mass 800 kg moving at a velocity of 30
  ms -1 is brought to rest by a braking force of
  1200 N.
• Calculate
• (a) its initial momentum
• (b) the time taken to stop the car.

• (a) p mv
• 800 kg x 30 ms-1
• momentum 24 000 kg ms-1
• (b) F ?(mv) / ?t
• 1200N 24 000 kg ms-1 / ?t
• ?t 24 000 kg ms-1 / 1200N
• time 20 seconds

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Question 2

• A car of mass 750kg travelling at a speed of
  4.0ms -1 is struck from behind by another
  vehicle. The impact lasts for 0.30s and causes
  the speed of the car to increase to 6.0ms -1 .
• Calculate
• (a) the change in momentum of the car due to the
  impact.
• (b) the impact force.

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Question 2

• A car of mass 750kg travelling at a speed of
  4.0ms -1 is struck from behind by another
  vehicle. The impact lasts for 0.30s and causes
  the speed of the car to increase to 6.0ms -1 .
• Calculate
• (a) the change in momentum of the car due to the
  impact.
• (b) the impact force.

• (a) ?p ?mv
• mass is constant, so
• ?p m ?v
• 750 kg x (6.0 4.0) ms-1
• momentum change 1 500 kg ms-1

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Question 2

• A car of mass 750kg travelling at a speed of
  4.0ms -1 is struck from behind by another
  vehicle. The impact lasts for 0.30s and causes
  the speed of the car to increase to 6.0ms -1 .
• Calculate
• (a) the change in momentum of the car due to the
  impact.
• (b) the impact force.

• (a) ?p ?mv
• mass is constant, so
• ?p m ?v
• 750 kg x (6.0 4.0) ms-1
• momentum change 1 500 kg ms-1
• (b) F ?(mv) / ?t
• 1 500 kg ms-1 / 0.30s
• force 5000 N

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Impulse, ?p

• Impulse is equal to the change of momentum
  produced by a force over a period of time.
• Impulse, ?p F?t ?(mv)
• ?p is measured in newton seconds (Ns)

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Impulse, ?p

• Impulse is equal to the change of momentum
  produced by a force over a period of time.
• Impulse, ?p F?t ?(mv)
• ?p is measured in newton seconds (Ns)

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Impulse, ?p

• Impulse is equal to the change of momentum
  produced by a force over a period of time.
• Impulse, ?p F?t ?(mv)
• ?p is measured in newton seconds (Ns)

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Impulse caused by a golf club(Breithaupt page 8)
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Force time graphs(Breithaupt page 6)

• Impulse is equal to the area under a force-time
  graph.

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Calculation Example(Breithaupt page 9)
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Calculation Example(Breithaupt page 9)
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Calculation Example(Breithaupt page 9)
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Graph Question

• Calculate the impulse and change in velocity
  caused to mass of 6kg from the graph opposite.

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Graph Question

• Calculate the impulse and change in velocity
  caused to mass of 6kg from the graph opposite.
• Area impulse
• 3N x (5 - 2 )s
• impulse 9 Ns

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Graph Question

• Calculate the impulse and change in velocity
  caused to mass of 6kg from the graph opposite.
• Area impulse
• 3N x (5 - 2 )s
• impulse 9 Ns
• ?(mv) 6kg x ?(v)
• therefore, ?(v) 9 / 6
• velocity change 1.5 ms-1

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Conservation of Linear Momentum

• The total linear momentum of an isolated system
  of bodies remains constant
• An isolated system is one where no external
  forces (e.g. friction or air resistance) acts on
  the interacting bodies.

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Conservation of Linear Momentum

• The total linear momentum of an isolated system
  of bodies remains constant
• An isolated system is one where no external
  forces (e.g. friction or air resistance) acts on
  the interacting bodies.

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Question

• A trolley of mass 4kg moving at 5ms-1 collides
  with another initially stationary trolley of mass
  3kg. If after the collision the trolleys move off
  attached together calculate their common final
  velocity.

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Question

• A trolley of mass 4kg moving at 5ms-1 collides
  with another initially stationary trolley of mass
  3kg. If after the collision the trolleys move off
  attached together calculate their common final
  velocity.
• Initial total linear momentum of the system
• momentum of 4kg trolley momentum of 3kg
  trolley
• (4kg x 5ms-1) (3kg x 0ms-1)
• 20 kgms-1

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Question

• A trolley of mass 4kg moving at 5ms-1 collides
  with another initially stationary trolley of mass
  3kg. If after the collision the trolleys move off
  attached together calculate their common final
  velocity.
• Initial total linear momentum of the system
• momentum of 4kg trolley momentum of 3kg
  trolley
• (4kg x 5ms-1) (3kg x 0ms-1)
• 20 kgms-1

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• Conservation of linear momentum
• Final total linear momentum of the system
• must also 20 kgms-1
• (total mass x final common velocity) 20 kgms-1
• (4kg 3kg) x v 20 kgms-1
• 7v 20
• v 20 / 7
• Final common velocity 2.86 ms-1

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• Conservation of linear momentum
• Final total linear momentum of the system
• must also 20 kgms-1
• (total mass x final common velocity) 20 kgms-1
• (4kg 3kg) x v 20 kgms-1
• 7v 20
• v 20 / 7
• Final common velocity 2.86 ms-1

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• Conservation of linear momentum
• Final total linear momentum of the system
• must also 20 kgms-1
• (total mass x final common velocity) 20 kgms-1
• (4kg 3kg) x v 20 kgms-1
• 7v 20
• v 20 / 7
• Final common velocity 2.86 ms-1

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Elastic and inelastic collisions

• ELASTIC KINETIC energy is conserved
• INELASTIC Some (or all) KINETIC energy is
  transformed into thermal or other forms of
  energy.
• In both types of collision both the total energy
  and momentum are conserved

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Elastic and inelastic collisions

• ELASTIC KINETIC energy is conserved
• INELASTIC Some (or all) KINETIC energy is
  transformed into thermal or other forms of
  energy.
• In both types of collision both the total energy
  and momentum are conserved

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Elastic and inelastic collisions

• ELASTIC KINETIC energy is conserved
• INELASTIC Some (or all) KINETIC energy is
  transformed into thermal or other forms of
  energy.
• In both types of collision both the total energy
  and momentum are conserved

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Elastic and inelastic collisions

• ELASTIC KINETIC energy is conserved
• INELASTIC Some (or all) KINETIC energy is
  transformed into thermal or other forms of
  energy.
• In both types of collision both the total energy
  and momentum are conserved

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Collision question continued

• Was the collision in the previous example elastic
  or inelastic?

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Collision question continued

• Was the collision in the previous example elastic
  or inelastic?
• Kinetic energy ½ x mass x (speed)2
• Total initial KE KE of 4kg trolley
• ½ x 4kg x (5 ms-1)2 2 x 25 50 J
• Total final KE KE of combined 7kg trolley
• ½ x 7kg x (2.86 ms-1)2 3.5 x 8.18 28.6 J
• Kinetic energy reduced Collision INELASTIC

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Collision question continued

• Was the collision in the previous example elastic
  or inelastic?
• Kinetic energy ½ x mass x (speed)2
• Total initial KE KE of 4kg trolley
• ½ x 4kg x (5 ms-1)2 2 x 25 50 J
• Total final KE KE of combined 7kg trolley
• ½ x 7kg x (2.86 ms-1)2 3.5 x 8.18 28.6 J
• Kinetic energy reduced Collision INELASTIC

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Collision question continued

• Was the collision in the previous example elastic
  or inelastic?
• Kinetic energy ½ x mass x (speed)2
• Total initial KE KE of 4kg trolley
• ½ x 4kg x (5 ms-1)2 2 x 25 50 J
• Total final KE KE of combined 7kg trolley
• ½ x 7kg x (2.86 ms-1)2 3.5 x 8.18 28.6 J
• Kinetic energy reduced Collision INELASTIC

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Collision question continued

• Was the collision in the previous example elastic
  or inelastic?
• Kinetic energy ½ x mass x (speed)2
• Total initial KE KE of 4kg trolley
• ½ x 4kg x (5 ms-1)2 2 x 25 50 J
• Total final KE KE of combined 7kg trolley
• ½ x 7kg x (2.86 ms-1)2 3.5 x 8.18 28.6 J
• Kinetic energy reduced Collision INELASTIC

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Explosions

• KINETIC energy is increased
• Both the total energy and momentum are conserved

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Question

• A gun of mass 3kg fires a bullet of mass 15g. If
  the bullet moves off at a speed of 250ms-1
  calculate the recoil speed of the gun.
• Conservation of linear momentum Final total
  linear momentum of the system must also 0
  kgms-1

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Question

• A gun of mass 3kg fires a bullet of mass 15g. If
  the bullet moves off at a speed of 250ms-1
  calculate the recoil speed of the gun.
• Initial total linear momentum of the system
• momentum of the gun momentum of the bullet
• (3kg x 0ms-1) (15g x 0ms-1)
• 0 kgms-1
• Conservation of linear momentum Final total
  linear momentum of the system must also 0 kgms-1

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Question

• A gun of mass 3kg fires a bullet of mass 15g. If
  the bullet moves off at a speed of 250ms-1
  calculate the recoil speed of the gun.
• Initial total linear momentum of the system
• momentum of the gun momentum of the bullet
• (3kg x 0ms-1) (15g x 0ms-1)
• 0 kgms-1
• Conservation of linear momentum Final total
  linear momentum of the system must also 0 kgms-1

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Question

• A gun of mass 3kg fires a bullet of mass 15g. If
  the bullet moves off at a speed of 250ms-1
  calculate the recoil speed of the gun.
• Initial total linear momentum of the system
• momentum of the gun momentum of the bullet
• (3kg x 0ms-1) (15g x 0ms-1)
• 0 kgms-1
• Conservation of linear momentum Final total
  linear momentum of the system must also 0 kgms-1

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• Therefore
• (bullet mass x velocity) (gun mass x velocity)
  0
• (0.015kg x 250ms-1) (3kg x gun velocity) 0
• (3.75) (3 x gun velocity) 0
• 3 x gun velocity - 3.75
• gun velocity - 3.75 / 3 - 1.25 ms-1
• The MINUS sign indicates that the guns velocity
  is in the opposite direction to that of the
  bullet
• Gun recoil speed 1.25 ms-1

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• Therefore
• (bullet mass x velocity) (gun mass x velocity)
  0
• (0.015kg x 250ms-1) (3kg x gun velocity) 0
• (3.75) (3 x gun velocity) 0
• 3 x gun velocity - 3.75
• gun velocity - 3.75 / 3 - 1.25 ms-1
• The MINUS sign indicates that the guns velocity
  is in the opposite direction to that of the
  bullet
• Gun recoil speed 1.25 ms-1

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• Therefore
• (bullet mass x velocity) (gun mass x velocity)
  0
• (0.015kg x 250ms-1) (3kg x gun velocity) 0
• (3.75) (3 x gun velocity) 0
• 3 x gun velocity - 3.75
• gun velocity - 3.75 / 3 - 1.25 ms-1
• The MINUS sign indicates that the guns velocity
  is in the opposite direction to that of the
  bullet
• Gun recoil speed 1.25 ms-1

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• Therefore
• (bullet mass x velocity) (gun mass x velocity)
  0
• (0.015kg x 250ms-1) (3kg x gun velocity) 0
• (3.75) (3 x gun velocity) 0
• 3 x gun velocity - 3.75
• gun velocity - 3.75 / 3 - 1.25 ms-1
• The MINUS sign indicates that the guns velocity
  is in the opposite direction to that of the
  bullet
• Gun recoil speed 1.25 ms-1

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Internet Links

• Effect of impulse - NTNU
• Collisions along a straight line - NTNU
• 1D collision showing momentum and ke - NTNU
• 2D collisions - NTNU
• 2D Collisions - Explore Science
• Two dimensional collisions - Virginia
• Elastic Inelastic Collisions - Fendt
• Newton's Cradle - Fendt
• Gaussian gun - NTNU
• Dropping a load onto a trolley - momentum -
  netfirms
• Ballistic Pendulum - NTNU

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Core Notes from Breithaupt pages 4 to 17

• Define what is meant by momentum. State its unit.
• Explain how force is related to the rate of
  change of momentum.
• What is meant by impulse? How can impulse be
  found graphically? Copy Figure 3 on page 6.
• State the principle of conservation of momentum.
• Redo the worked example on page 13 this time with
  the first rail wagon moving at 4ms-1 colliding
  with another now of mass 2000kg.
• Define what is meant by (a) an elastic and (b) an
  inelastic collision.
• Redo the worked example on page 15 this time with
  the first rail wagon moving at 4ms-1 colliding
  with another now of mass 6000kg.
• Explain how the principle of conservation of
  momentum applies in an explosion.
• Redo summary question 1 on page 17 this time with
  a shell of mass 3.0kg being fired from a gun of
  mass 1000kg.

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Notes from Breithaupt pages 4 to 7Force
Momentum

• Define what is meant by momentum. State its unit.
• Explain how force is related to the rate of
  change of momentum.
• What is meant by impulse? How can impulse be
  found graphically? Copy Figure 3 on page 6.
• Show how the version of Newtons 2nd law of
  motion on page 5 can be used to derive the
  equation F ma
• Redo the worked example on page 7 this time with
  a force of 20N on a mass of 200kg.
• Try the summary questions on page 7

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Notes from Breithaupt pages 8 to 10Impact Forces

• Redo the worked example on page 8 this time with
  a velocity increase of 25ms-1 over a time of
  20ms.
• Explain how the relationship between force and
  momentum change is relevant to vehicle safety.
• Explain why a greater force is needed to send a
  ball back along its initial path than to deflect
  it at an angle.
• Redo the worked example on page 10 this time with
  a ball of mass 0.30kg moving at an initial speed
  of 15ms-1.
• Try the summary questions on page 10

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Notes from Breithaupt pages 11 to 13Conservation
of momentum

• State the principle of conservation of momentum.
• Redo the worked example on page 13 this time with
  the first rail wagon moving at 4ms-1 colliding
  with another now of mass 2000kg.
• Show how the version of Newtons 3rd law of
  motion on page 11 can be used to derive the
  principle of conservation of momentum.
• Explain how the principle of conservation of
  momentum can be verified experimentally.
• Explain how the principle of conservation of
  momentum applies in a head-on collision.
• Try the summary questions on page 13

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Notes from Breithaupt pages 14 15Elastic and
inelastic collisions

• Define what is meant by (a) an elastic and (b) an
  inelastic collision.
• Redo the worked example on page 15 this time with
  the first rail wagon moving at 4ms-1 colliding
  with another now of mass 6000kg.
• What is a totally inelastic collision?
• Explain how conservation of energy can still
  apply to an inelastic collision. What else is
  conserved in this type of collision?
• Try the summary questions on page 15

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Notes from Breithaupt pages 16 17Explosions

• Explain how the principle of conservation of
  momentum applies in an explosion.
• Redo summary question 1 on page 17 this time with
  a shell of mass 3.0kg being fired from a gun of
  mass 1000kg.
• Explain how the principle of conservation of
  momentum can be verified experimentally with an
  explosive interaction.
• Try the other summary questions on page 17