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Year  11  Physics  Unit  2  

Area  of  Study  2:  Motion  

Chapter  6  –  Momentum  and  Energy  

 Outcome:  On  completion  of  this  area  of  study,  you  should  be  able  to  investigate,  analyse  and  mathematically  model  motion  of  particles  and  bodies  in  terms  of  Aristotelian,  Galilean  and  Newtonian  theories.  By  the  end  of  this  Chapter:  You  will  have  covered  material  from  the  study  of  movement  including:  •  Momentum  and  impulse    •  Work  done  as  a  change  in  energy    •  Hooke’s  law    •  Kinetic,  gravitational  and  elastic  potential  energy    •  Energy  transfers  •  Power    Momentum  and  Impulse  How  difficult  is  it  to  stop  a  moving  object?  How  difficult  is  it  to  make  a  stationary  object  move?  The  answer  to  both  of  these  questions  depends  on  two  physical  characteristics  of  the  object:  • The  objects  mass    • How  fast  the  object  was  moving,  or  how  fast  you  want  it  to  move.  The  product  of  these  two  physical  characteristics  is  called  momentum.  The  momentum,  p,  of  an  object  of  mass,  m,  with  a  velocity,  v,  is  defined  as:    p  =  mv.  Momentum  is  a  vector  quantity  and  has  SI  units  of  kg  m  s-­‐1.    

 Example  What  is  the  momentum  of  a  train  of  mass  8.0  ×  106  kg  that  is  travelling  at  a  speed  of  15  m  s-­‐1  in  a  northerly  direction?  Solution:  

m  =  8.0  ×  106  kg,  v  =  15  m  s-­‐1  north  p  =  mv  

           =  8.0  ×  106  kg  ×  15  m  s-­‐1  north            =  1.2  ×  107  kg  m  s-­‐1  north  

Making   an   object   stop,   or   causing   it   to   start   moving,   requires   a   non-­‐zero   net   force.   The  relationship  between   the  net   force  applied   to  an  object  and   its  momentum  can  be  explored  by  applying  Newtons  second  law  to  the  object.  

𝑭𝒏𝒆𝒕 = 𝑚𝒂  

𝑭𝒏𝒆𝒕 = 𝑚∆𝑣∆t  

𝑭𝒏𝒆𝒕∆𝒕 = 𝑚∆𝒗  

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The  product  Fnet∆t  is  called  the  impulse  of  the  net  force.  The  impulse  of  any  force  is  defined  as  the  product  of  the  force  and  the  time  interval  over  which  it  acts.  Impulse  is  a  vector  quantity  with  SI  units  of  N  s.  

𝑚∆𝑣 = 𝑚 𝑣 − 𝑢 = 𝑚𝑣 −𝑚𝑢 = 𝑝! − 𝑝!  pf  =  the  final  momentum  of  the  object  pi  =  the  initial  momentum  of  the  object.    Thus,  the  effect  of  a  net  force  on  the  motion  of  an  object  can  be  summarised  by  the  statement:    

impulse  =  change  in  momentum.    Example  

A  30  g  squash  ball  hits  a  wall  horizontally  at  a  speed  of  15  m  s-­‐1  and  bounces  back  in  the  opposite  

direction  at  a  speed  of  12  m  s-­‐1.  It  is  in  contact  with  the  wall  for  an  interval  of  1.5  ×  10-­‐3  seconds.    (a)  What  is  the  change  in  momentum  of  the  squash  ball?  (b)  What  is  the  impulse  on  the  squash  ball?    (c)  What  is  the  magnitude  of  the  force  exerted  by  the  wall  on  the  squash  ball?  Solution:  Assign  the  initial  direction  of  the  ball  as  positive.  (a)  Δp=mv-­‐mu  =  m(v  -­‐  u)  

                       =  0.030  kg  (-­‐12  m  s-­‐1  -­‐  15  m  s-­‐1)  =  0.030  kg  ×  -­‐27  m  s-­‐1  =  -­‐0.81  kg  m  s-­‐1  

 (b)  Impulse  of  the  net  force  on  the  squash  ball  =  change  in  momentum  of  the  squash  ball    

=  -­‐0.81  kg  m  s-­‐1  =  -­‐0.81  N  s  

It  can  be  shown  that  1  N  s  =  1  kg  m  s-­‐1.    We  know  that  the  net  force  on  the  squash  ball  is  the  force  exerted  by  the  wall  since  there  is  no  change  in  the  vertical  motion  of  the  ball.  It  is  reasonable  to  ignore  air  resistance.  

(c)  Magnitude  of  impulse  =  FΔt  ⇒  0.81  N  s  =  F  ×  1.5  ×  10-­‐3  s  ⇒F=  0.81Ns/1.5×10-­‐3  s  ⇒  F  =  540  Ν  

Impulse  from  a  graph  The  force  that  was  determined  in  sample  problem  6.10  was  actually  the  average  force  on  the  squash  ball.  In  fact,  the  force  acting  on  the  squash  ball  changes,  reaching  its  maximum  magnitude  when  the  centre  of  the  squash  ball  is  at  its  smallest  distance  from  the  wall.  The  impulse  (I)  delivered  by  a  changing  force  is  given  by:  

I  =  Fav  Δt.  

If  a  graph  of  force  versus  time  is  available,  the  impulse  can  be  determined  from  the  area  under  the  graph.  (You  might  recall  that  the  displacement  of  an  object  can  be  determined  by  calculating  the  area  under  its  velocity-­‐versus-­‐  time  graph  and  displacement  =  vav  Δt.  Similarly,  the  change  in  velocity  of  an  object  can  be  determined  by  calculating  the  area  under  its  acceleration-­‐versus-­‐time  graph  and  change  in  velocity  =  aav  Δt.)  

Example:  The  graph  below  describes  the  changing  horizontal  force  on  a  40  kg  ice-­‐skater  as  she  begins  to  move  from  rest.  Estimate  her  speed  after  2.0  seconds.  

The  magnitude  of  the  impulse  on  the  skater  can  be  determined  by  calculating  the  area  under  the  graph.  

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This  can  be  determined  either  by  counting  squares  or  by  finding  the  shaded  area.  

Magnitude  of  impulse  =  area  A  +  area  B  +  area  C    

=12×1.1𝑠×400𝑁 + 0.9𝑠×200𝑁 +

12×0.9𝑠×200𝑁  

=  220  N  s  +  180  N  s  +  90  N  s  =  490  N  s  

Impulse  =  change  in  momentum  =  mΔv    

⇒  490  N  s  =  40  kg  ×  Δv  

⇒                      Δv  =  490Ns  /40  kg=  12  m  s-­‐1      As  her  initial  speed  is  zero  (she  started  from  rest),  her  speed  after  2.0  seconds  is  12  m  s-­‐1.    Airbags  The  extent  of  injuries  during  a  collision  is  not  only  dependent  on  the  size  of  the  force  but  also  the  duration  and  deflection  resulting  from  the  applied  force.  An   increase   in   localised  pressure  will  result   in  a  greater  compression   or   deflection   of   the   skull.   The   air   bag   reduces   the    localised   pressure   by   increasing   the  contact  surface  area  and  decreasing  the  force.  

An  air  bag  has  a  contact  surface  area  of  about  0.2  m2  compared  with  0.05  m2  for  a  seat  belt.  This  reduces  injuries   caused   by   seat   belts,   such   as   bruising   and   broken   ribs   and   collar   bones,   since   it   increases   the  stopping   time.   It   also   supports   the   head   and   chest,   preventing   high   neck   loads   caused   by   the   seat   belt  restraining   the  upper   torso.  Most   importantly,   it  prevents   the  high   forces  caused  by  contact  of   the  head  with  the  steering  wheel.  

 

   

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6.2  Conservation  of  momentum    The  most  significant  feature  of  momentum  is  that  it  is  conserved.  This  means  that  the  total  momentum  in  any  complete  system  will  be  constant.  For  this  reason  momentum  is  very  useful  in  investigating  the  forces  experienced  by  two  colliding  objects—as  long  as  they  are  unaffected  by  outside  forces.  The  law  of  conservation  of  momentum,  as  it  is  known,  is  derived  from  Newton’s  third  law  (see  textbook)  and  is  given  as:  

m1u1  m2u2    =  m1v1  m2v2    

In  other  words,  the  total  momentum  before  colliding  is  the  same  as  the  total  momentum  after  the  collision.  

     

It  is  most  important  to  realise  that  momentum  is  only  conserved  in  an  isolated  system;  that  is,  a  system  in  which  no  external  forces  affect  the  objects  involved.  The  only  forces  involved  are  the  action–reaction  forces  on  the  objects  in  the  collision.  

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Modelling  real  collisions  The  Law  of  Conservation  of  Momentum  makes  it  possible  to  predict  the  con-­‐  sequences  of  collisions  between  two  cars  or  between  two  people  on  a  sporting  field.  For  example,  if  a  2000  kg  delivery  van  travelling  at  30  m  s-­‐1  (108  km  h-­‐1)  collided  with  a  small,  stationary  car  of  mass  1000  kg,  the  speed  of  the  tangled  wreck  (the  two  vehicles  locked  together)  could  be  predicted.  However,  you  would  need  to  assume  that  the  frictional  forces  and  driving  force  acting  on  both  cars  were  zero  after  the  collision.  A  reasonably  good  estimate  can  be  made  of  the  speed  of  the  tangled  wreck  immediately  after  the  collision  in  this  way.  The  initial  momentum  pi  of  the  system  is  given  by:    

 pvan  +  pcar  =    

 

where  the  initial  direction  of  the  van  is  taken  to  be  positive.  The  momentum  of  the  system  after  the  collision  pf  is  the  momentum  of  just  one  object  the  tangled  wreck.    

pf  =           where  v  is  the  velocity  of  the  tangled  wreck  after  the  collision.    

 

But  since  pf  =  pi,  

 

   

The  speed  of  the  small  car  changes  a  lot  more  than  the  speed  of  the  large  van.  However,  the  change  in  the  momentum  of  the  car  is  equal  and  opposite  to  that  of  the  van.  

Δpcar  =  mΔv=    

Δpvan  =  mΔv=    

 

If  the  small  car  hit  the  stationary  van  at  a  speed  of  30  m  s-­‐1,  and  the  two  vehicles  locked  together,  the  speed  of  the  tangled  wreck  would  be  less  than  20  m  s-­‐1.  Apply  the  Law  of  Conservation  of  Momentum  to  predict  the  speed  of  the  tangled  wreck  immediately  after  the  collision.  

   

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6.3  Work  In  physics  work  is  done  on  an  object  by  the  action  of  a  force  or  forces.  The  object  is  often  referred  to  as  the  load.  Many  interactions  are  complex  and  there  is  often  more  than  one  force  present.  As  work  can  only  be  done   in   the  presence  of   a   force,   it   is   imperative   that   any   time   the  work  done   in  a  particular   situation   is  being  discussed,  the  relevant  force,  forces  or  net  force  should  be  clearly  stated.  For  clarity,  the  item  upon  which  the  work  is  done,  the  load,  should  also  be  specified.  Clearly  specified  examples  of  work  are:  

• The  work  done  by  gravity  on  a  diver  as  she  falls    • The  work  done  by  arm  muscles  on  a  school  bag  lifted  to  your  shoulder    • The  work  done  by  the  heart  muscle  on  a  volume  of  blood  during  a    contraction    • The  work  done  by  the  net  force  acting  on  a  cyclist  climbing  a  hill.      

Always  being  clear  about  the  particular  forces  and  objects  examined  will  prevent  considerable  confusion  in  this  area  of  study.    For  work  to  be  done  on  a  body,  the  energy  of  the  body  must  change.  Thus  the  work  done  is  measured  in  joules,  which  is  also  the  unit  of  energy.    

 

Work  done  by  a  constant  force  If  the  net  force  acting  on  an  object  in  a  particular  situation  has  a  constant  value,  or  if  it  is  appropriate  to  utilise  an  average  force  value,  then:    

 

   

 

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 Work  and  friction  If  an  object  is  forced  to  move  across  a  surface  by  the  application  of  a  force,   its  motion  may  be  slowed  by   friction.   In   this  case   the  applied  force   is   doing   work   on   the   object   and   the   frictional   force   can   be  considered  to  be  doing  ‘negative’  work  on  the  object.  

Worked  example  6.3A  Calculate  the  work  done  against  gravity  by  an  athlete  of  mass  60  kg  competing  in  the  Great  rialto  Stair  trek  illustrated  in  Figure  6.16.  Use  g  =  9.8  m  s−2.  Assume  the  athlete  climbs  at  a  constant  speed.  

   

   

 

   

       Upward  force  does  no  work  A  more  difficult  idea  to  comprehend  is  that  in  the  apparent  absence  of  friction,  a  force  can  be  exerted  on  an  object  yet  do  no  work  on  it.  For  example,  when  a  person  carries  an  armload  of  books  horizontally  the  upward  force  does  no  work  on  the  books  since  the  direction  of  the  applied  force  (i.e.  up)  is  at  right  angles  to  the  displacement  (i.e.  horizontal).  Similarly  if  a  person  is  holding  a  heavy  item,  such  as  a  TV,  stationary,  they  may  be  exerting  great  effort  yet  do  no  work  on  it.  

 

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Force–displacement  graphs  A  graphical  approach  can  also  be  used  to  understand  the  action  of  a  force  and  the  work  expended  in  the  direction   of   motion.   This   is   particularly   useful   in   situations   in   which   the   force   is   changing   with  displacement.  The  area  under  a  graph  of  force  against  displacement  always  represents  the  work  produced  by  the  force  

 Impulse  and  work  The  concepts  of  impulse  and  work  seem  quite  similar  and,  when  solving  problems,  can  easily  be  confused.  Actually,  problems  focusing  on  forces  in  collisions  may  be  solved  using  either  concept,  but  it  should  be  understood  that  each  is  derived  from  a  different  idea.  Impulse  comes  from  an  understanding  of  the  action  of  a  force  on  an  object  over  time  and  is  equal  to  the  change  in  momentum  the  force  produces.  Work  is  related  to  the  action  of  a  force  on  an  object  as  it  moves  the  object,  or  part  of  it,  through  some  displacement.  This  equals  the  change  in  the  object’s  energy,  Δ  E.  

Summarising:    

•  Impulse  is  equal  to  F  ×  Δ  t,  is  equivalent  to  Dp,  has  the  units  newton  seconds  (N  s),  and  can  be  determined  from  the  area  under  a  force–time  graph.  

•  Work  is  equal  to  F×x,  is  equivalent  to  Δ  E,  has  the  units  joules  (J),  and  can  be  determined  from  the  area  under  a  force–displacement  graph.      

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6.4  Mechanical  Energy  Kinetic energy

We observe many different forms of energy each day. We have come to take for granted the availability of light, heat, sound and electrical energy whenever we require it. We rely upon the chemical potential energies that re available when petrol, diesel and LPG are burnt to run our vehicles, and food to fuel our bodies. Whenever work is done, energy is expended.

 If a moving object of mass, m, and initial velocity, u, experiences a constant net force, F, for time, t, then a uniform acceleration results. The velocity will increase to a final value, v. Work will have been done during the time the force is applied. Since work is equivalent to the change in kinetic energy of the object, there should be a relationship linking the two quantities. This can be found from the definition for work when the net applied force is in the direction of the displacement:

W = ΣFx �

Now substituting Newton’s second law F = ma we get:

W = max �

Using one of the earlier equations of motion: v2 = u2 + 2ax

Substituting and rearranging gives: W = ½ mv2 − ½ mu2 � but W = ΔE.

�If it is accepted that the work done results in a change in kinetic energy then an object of mass m with a speed v has kinetic energy equal to ½ mv2.

 Like all forms of energy, kinetic energy is a scalar quantity and is measured in joules (J). There is no direction associated with it. The kinetic energy of an object depends solely on its mass and velocity. The approximate kinetic energy of various moving objects is given in Table 6.3.

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Potential  energy  

An  object  can  have  energy  not  only  because  of  its  motion,  but  also  as  a  result  of  its  shape  or  position.  This  is   called  potential   energy.   A   gymnast,   crouched   ready   to   jump,  has  potential   energy.  During   the   jump,  work   is  being  done  by   the   force  exerted  by   the  gymnast,   and  potential   energy   is   converted   into  kinetic  energy  from  the  stores  of  chemical  energy  in  the  muscles  of  the  gymnast’s  body.  

Gravitational  potential  energy  

An  athlete  at  the  top  of  a  high-­‐jump  has  gravitational  potential  energy  because  of  his  position.  As  he  falls,  work  is  done  (Figure  6.20).  Recall  that  in  this  case  the  work  done  is  given  by:  

Work done = ΣFx �The force acting on the body is simply the force due to gravity also called the person’s weight:

Weight = mg �

The displacement that occurs is in a vertical direction and can be described as a change in height, Δh. Replacing F and x with these equivalent terms gives:

W = mgΔh � Similarly, the work done in raising the athlete against a gravitational field is stored as gravitational potential energy; hence, the athlete has a change in potential energy:

ΔUg = mgΔh

 

     

   

 

   

 

 

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Elastic  materials  and  elastic  potential  energy  The  third  aspect  of  mechanical  energy  that  we  will  study  is  elastic  potential   energy.  Like  gravitational  potential  energy  it  occurs  in  situations  where  energy  can  be  considered  to  be  stored  temporarily  so  that,  when  this  energy   is  released,  work  may  be  done  on  an  object.  Elastic  potential  energy   is  stored  when  a  spring   is   stretched,   a   rubber  ball   is   squeezed,   air   is   compressed   in   a   tyre   or   a   bungee   jumper’s   rope   is  extended  during  a  fall.  

Ideal  springs  obey  Hooke’s  law    Springs   are   very   useful   items   in   our   everyday   life   due   to   the   consistent   way   in   which   many   of   them  respond  to  forces  and  store  energy.  When  a  spring  is  stretched  or  compressed  by  an  applied  force  we  say  that  elastic  potential  energy  is  being  stored.    

In  order  to  store  this  energy  work  must  be  done  on  the  spring.    

Recall  W  =  Fx.  This  formula  can  therefore  be  used  when  a  set  force,  F,  has  been  applied  to  a  spring  and  a  given  compression  or  extension,  Δx,  occurs.  However,  we  are  usually  interested  in  examining  how  a  spring  will  behave  in  a  range  of  conditions.    

Consider a spring stretched by the application of a steadily increasing force. As the force increases, the extension of the spring, Δx, can be graphed against the applied force, F. Imagine a spring hanging vertically and gradually adding more and more weight to it so that is stretches, a well-designed springs, will extend in proportion to the applied force.

For example, if a 10 N force produced an extension of 6 cm, then a 20 N force would produce an extension of 12 cm. These items are called ideal springs. The resulting graph of applied force versus extension would be linear as in Figure 6.22.

• The gradient of this graph tells us the force (N), required to produce each unit of extension.

• The gradient of the graph is called the spring constant, k, measured in N m−1.

• The gradient is the stiffness of the spring, and for an ideal spring this gradient has a set value (i.e. the F vs Δx graph is a straight line).

• A very stiff spring that is difficult to stretch would have a steep gradient; that is, a large value of k.

• k is sometimes called the stiffness constant or force constant of a spring.

A spring constant of k = 1500 N m−1 indicates that for every metre that the spring is stretched or compressed, a force of 1500 N is required. This does not necessarily mean that the spring can be stretched by 1 m, but it tells us that the force and the change in length are in this proportion.

The relationship between the applied force and the subsequent extension or compression of an ideal spring is known as Hooke’s law.

   

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Calculating elastic potential energy

Work must be done in order to store elastic potential energy in any elastic material. Essentially the energy is stored within the atomic bonds of the material as it is compressed or stretched. The amount of elastic potential energy stored is given by the area under the force–extension graph for the item.

For materials that obey Hooke’s law an expression can be derived for the area under the F v. x graph. Work done = area under F–x graph = area of a triangle = ½ ⋅ base ⋅ height = ½ ⋅ Δ x ⋅ F , But since F = kΔx: Workdone = ½ ⋅Δx⋅ (kΔx) W = ½ kΔx 2 = the elastic potential energy stored during the extension/compression Although many materials (at least for a small load) extend in proportion to the applied force, many materials have force–extension graphs more like that shown in Figure 6.23. For these materials the area under the F–Δx graph must be used to determine the elastic potential energy stored.