Physics Chapter 6: Momentum and Collisions

 Force is Not Always Constant  Application of Force May Vary with Time

Momentum and Collisions  Impulse  The Product of the Average Force and the Time Interval During Which the Force Acts  A Change in Momentum

Momentum and Collisions  Impulse  Units  Force = Newton  Time = Second  Impulse = Newton * Second (N*s)

Momentum and Collisions  Linear Momentum (p)  The Reaction of an Object to a Force Exerted Upon It  The Product of an Object’s Mass and Velocity

Momentum and Collisions  Linear Momentum (p)  Vector Quantity  Direction is Always the Same as the Direction of the Velocity Upon the Object

Momentum and Collisions  Linear Momentum (p)  Units  Mass = kilogram  Velocity = meters / second  Linear Momentum = kilogram*meter/second  (kg*m/s)

Momentum and Collisions  Impulse – Momentum Theorem

Momentum and Collisions  Impulse – Momentum Theorem

Momentum and Collisions  Impulse – Momentum Theorem  Impulse = Final Momentum – Initial Momentum  Impulse = Change in Momentum

Momentum and Collisions  Impulse – Momentum Theorem  Impulse Time is Usually Very Short  Experimental Measurement of Impulse Force is Usually Difficult  I – M Theorem Allows Calculation of Impulse and Subsequently the Force Applied

Momentum and Collisions  Impulse – Momentum Theorem  The Impulse Produced by a Net Force is Equal to the Change in the Object’s Momentum

Momentum and Collisions  Problem –Hilary strikes a 0.058kg golf ball with a force of 272N and gives it a velocity of 62.0m/s. How long was Hilary’s club in contact with the ball?

Momentum and Collisions  Solution  m = 0.058kg  F = 272N  v i = 0m/s  v f = 62m/s

Momentum and Collisions  Problem –A 6.0g bullet is fired at a velocity of 350m/s into a container of ballistic gelatin that stops the bullet in 1.8ms. What is the average force that stops the bullet?

Momentum and Collisions  Solution  m = 6g = 0.006kg   t = 1.8ms  v i = 350m/s  v f = 0m/s

Momentum and Collisions  Problem –A hockey puck has a mass of 0.115kg and is at rest. A hockey player makes a shot, exerting a constant force of 30.0N on the puck for 0.16s. With what speed does it head toward the goal?

Momentum and Collisions  Solution  m = 0.115kg  F = 30N   t = 0.16s

Momentum and Collisions  Problem –A 0.150kg ball, moving in the positive direction at 12m/s, is acted on by the impulse shown in the graph. What is the ball’s speed at 4.0 s?

Momentum and Collisions  Solution  m = 0.150kg  v = 12m/s   t = 4.0s

Momentum and Collisions  Problem –Small rockets are used to make tiny adjustments in the speeds of satellites. One such rocket has a thrust of 35N. If it is fired to change the velocity of a 72,000kg spacecraft by 63cm/s, how long should it be fired?

Momentum and Collisions  Solution  F = 35N  m = 7.2x10 4 kg   v = 63cm/s

Momentum and Collisions  Homework  Page  Problems  14 (160N Right)  16 (0.010s, 0.13m)  24 (a, 2.43m/s right b, 7.97x10 -2 m/s right)  26 (a, 2.5m/s right b, 3.0m/s right)

Momentum and Collisions  Conservation of Momentum  Work – Energy Theorem  The Work Done by a Net Force is Equal to the Change in the Object’s Kinetic Energy  Impulse – Momentum Theorem  The Impulse Produced by a Net Force is Equal to the Change in the Object’s Momentum

Momentum and Collisions  Conservation of Momentum  Work – Energy Theorem  Conservation of Mechanical Energy  Impulse – Momentum Theorem  Conservation of Linear Momentum

Momentum and Collisions  Conservation of Linear Momentum  “System”  Internal Forces  Forces that Objects Within the System Exert on Each Other  External Forces  Forces Exerted on the Objects in the System by Agents Outside the System

Momentum and Collisions  Conservation of Linear Momentum  Collisions  Object A  Mass (m A )  Velocity (v A )  Weight (W A )  Object B  Mass (m B )  Velocity (v B )  Weight (W B )

Momentum and Collisions  Conservation of Linear Momentum  Collisions  Object A  Exerts Force on Object B (F A-B ) (Internal)  Weight  F g =mg (External)  Object B  Exerts Force on Object A (F B-A ) (Internal)  Weight  F g =mg (External)

Momentum and Collisions  Conservation of Linear Momentum  Collisions  Newton’s 3 rd Law Tells Us that These Forces are Equal and Opposite

Momentum and Collisions  Conservation of Linear Momentum  Collisions  In an Isolated System  No External Forces

Momentum and Collisions  Conservation of Linear Momentum  Collisions  The Total Linear Momentum of an Isolated System Remains Constant. An Isolated System is One for Which the Vector Sum of the External Forces Acting on the System is Zero

Momentum and Collisions  Collisions  Recoil  The momentum of a baseball changes when the external force of a bat is exerted on it. The baseball, therefore, is not an isolated system.  On the other hand, the total momentum of two colliding balls within an isolated system does not change because all forces are between the objects within the system.

Momentum and Collisions  Collisions  Propulsion in Space  Recoil Principle

Momentum and Collisions  Conservation of Linear Momentum  Two-Dimensional Collisions (Non-Linear)

Momentum and Collisions  Conservation of Linear Momentum  Two-Dimensional Collisions (Non-Linear)  Initial momentum of the moving ball is p Ci and the momentum of the stationary ball is zero.  The momentum of the system before the collision is equal to p Ci

Momentum and Collisions  Conservation of Linear Momentum  Two-Dimensional Collisions (Non-Linear)  After the collision, both billiard balls are moving and have momenta.  Ignoring friction with the tabletop, the system is closed.  Conservation of momentum can be used. The initial momentum equals the vector sum of the final momenta.

Momentum and Collisions  Conservation of Linear Momentum  Two-Dimensional Collisions (Non-Linear)  The sum of the components of the vectors before and after the collision must be equal.  Suppose the x-axis is defined to be in the direction of the initial momentum, then the y-component of the initial momentum is equal to zero.  Therefore, the sum of the final y-components also must be zero.

Momentum and Collisions  Conservation of Linear Momentum  Two-Dimensional Collisions (Non-Linear)  The y-components are equal in magnitude but are in the opposite direction and, thus, have opposite signs. The sum of the horizontal components also is equal.

Momentum and Collisions  Collisions  The Total Linear Momentum is Conserved When Two Objects Collide, Provided They Constitute an Isolated System

Momentum and Collisions  Collisions  Sometimes, Total KE of the Objects is Maintained  ex. Electrons  Sometimes, Total KE of the Objects is Transferred into Other Types of Energy  ex. Trains

Momentum and Collisions  Collisions  Elastic Collision  Total KE and Momentum of the System After the Collision is Equal to the Total KE and Momentum of the System Before the Collision  Inelastic Collision  Total KE and Momentum of the System After the Collision is NOT Equal to the Total KE and Momentum of the System Before the Collision

Momentum and Collisions  Collisions  Elastic Collision

Momentum and Collisions  Collisions  Examples (Elastic or Inelastic)  Pool Balls  Car Crash  Newtonian Swing  Photon  Punch  Basketball  Superball  Snowball

Momentum and Collisions  Collisions  Most Collisions are Neither Completely Elastic nor Completely Inelastic

Momentum and Collisions  Problem  Ben and Erika have a combined mass of 168kg. At an ice skating rink they stand at rest and Ben pushes Erika. Ben is sent in one direction at a velocity of 0.90m/s while Erika is sent the other direction at 1.2m/s. What is Ben’s mass?

Momentum and Collisions  Solution  m B = ?  m E = 168kg - m B  v E = -1.2m/s  v B = 0.90m/s

Momentum and Collisions  Problem  A car moving at 10.0m/s crashes into a barrier and stops in 0.050s. There is a 20.0kg child in the car. Assume that the child’s velocity is changed by the same amount as that of the car, and in the same time period. What is the impulse needed to stop the child?

Momentum and Collisions  Solution  v = 10.0m/s  t = 0.050s  m = 20.0kg

Momentum and Collisions  Problem  A car moving at 10.0m/s crashes into a barrier and stops in 0.050s. There is a 20.0kg child in the car. Assume that the child’s velocity is changed by the same amount as that of the car, and in the same time period. What is the average force on the child?

Momentum and Collisions  Solution  v = 10.0m/s  t = 0.050s  m = 20.0kg

Momentum and Collisions  Problem  Marble C, with mass 5.0g, moves at speed of 20.0cm/s. It collides with second marble, D, with mass 10.0g, moving 10.0cm/s in the same direction. After the collision, marble C continues with a speed of 8.0cm/s in the same direction. What are the marbles’ momenta before the collision?

Momentum and Collisions  Solution  m C = 5.0x10 -3 kg  v Ci = 0.20m/s  m D = 1.0x10 -2 kg  v Di = 0.10m/s  v Cf = 0.08m/s

Momentum and Collisions  Problem  Marble C, with mass 5.0g, moves at speed of 20.0cm/s. It collides with second marble, D, with mass 10.0g, moving 10.0cm/s in the same direction. After the collision, marble C continues with a speed of 8.0cm/s in the same direction. What is the momentum of marble C after the collision?

Momentum and Collisions  Solution  m C = 5.0x10 -3 kg  v Ci = 0.20m/s  m D = 1.0x10 -2 kg  v Di = 0.10m/s  v Cf = 0.08m/s

Momentum and Collisions  Problem  Marble C, with mass 5.0g, moves at speed of 20.0cm/s. It collides with second marble, D, with mass 10.0g, moving 10.0cm/s in the same direction. After the collision, marble C continues with a speed of 8.0cm/s in the same direction. What is the momentum of marble D after the collision?

Momentum and Collisions  Solution  m C = 5.0x10 -3 kg  v Ci = 0.20m/s  m D = 1.0x10 -2 kg  v Di = 0.10m/s  v Cf = 0.08m/s

Momentum and Collisions  Problem  Marble C, with mass 5.0g, moves at speed of 20.0cm/s. It collides with second marble, D, with mass 10.0g, moving 10.0cm/s in the same direction. After the collision, marble C continues with a speed of 8.0cm/s in the same direction. What is the speed of marble D after the collision?

Momentum and Collisions  Solution  m C = 5.0x10 -3 kg  v Ci = 0.20m/s  m D = 1.0x10 -2 kg  v Di = 0.10m/s  v Cf = 0.08m/s

Momentum and Collisions  Problem  A 2575kg van runs into the back of an 825kg compact car at rest. They move off together at 8.5m/s. Assuming that the friction with the road is negligible, calculate the initial speed of the van.

Momentum and Collisions  Solution  m V = 2575kg  m C = 825kg  v Vf = 8.5m/s  v Cf = 8.5m/s

Momentum and Collisions  Homework  Pages  Problems  32 (3.00m/s)  35 (a, 0.81m/s east b, 1.4x10 3 J)  37 (4.0m/s)  48 (14.5m/s north)