Further Mechanics 1 Unit Test 6 Elastic Collisions in Two Dimensions Mark Scheme

 

Mark scheme

Further Mechanics 1 Unit Test 6: Elastic collisions in two dimensions

Q

1

Scheme

Marks

AOs

Pearson Progression Step and Progress Descriptor

6th

By resolving and using the rule that the velocity of the ball after the collision parallel to the slope is unchanged,

M1

3.3

M1

3.3

M1

3.4

A1

1.1b

M1

1.1a

A1

1.1b

u cos 60 60  v co  s  

By resolving and using NEL formula finds an expression for the velocity of the ball after the collision perpendicular to the slope:

Use Newton's experimental law of restitution in the direction of  the impulse

60 0  v si n   eu sin 6

Uses KE formula and components to attempt to form an expression for the KE of the ball immediately after the collision with their  velocity  velocity components:

1 2

mv

2

 =

1

m

2

u

2

4

(1  3e2 )

Expression for KE of ball after collision fully correct as above.

 

Interprets given ratio to form equation: e quation: 7

e 

Solves:

2 3

12 4



1  3e 2

 

  

(6 marks) Notes

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1

 

Mark scheme

Further Mechanics 1 Unit Test 6: Elastic collisions in two dimensions

Q

2ai

Scheme

Marks

AOs

5th

Correctly applies Impulse formula to find Impulse vector, I: M1

 2   7     1  I  mv mu    0.2  2  10    2.4    Thus magnitude of Impulse

I

1.1a

A1

   (1) 2  (2.4)2   2.6 Ns

Pearson Progression Step and Progress Descriptor

A1

1.1b

Understand that impulse acts to  perpendicular the surface through the sphere's centre

(3)

5th

2aii 5   13 u    oe    12   13  States the unit vector for the Impulse as

B1

1.1b

Understand that impulse acts  perpendicular to the surface through the sphere's centre

(1) 2b

Chooses to use u.v to find the components of velocities  perpendicular to the wall. PI

M1

3.1b

Finds the component of the velocity of the sphere perpendicular to the wall before the collision: 5  7   13  155 oe   .  12    10     13  13

M1

1.1b

M1

1.1b

M1

1.1a

A1

1.1b

5th Understand that impulse acts  perpendicular to the surface through the sphere's centre

Likewise finds the component of the velocity of the sphere  perpendicular to the wall after the collision: 5  2   13   14 oe   .  12    2 13      13 

155

Using the NEL formula:

13 e  

Thus

e 

14 155

14

 

13

 

(5) (9 marks)

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2

 

Mark scheme

Further Mechanics 1 Unit Test 6: Elastic collisions in two dimensions

Notes

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3

 

Mark scheme

Further Mechanics 1 Unit Test 6: Elastic collisions in two dimensions

Q

Scheme

3a

Marks

AOs

Pearson Progression Step and Progress Descriptor

7th

 

 132  126

4sin =

65 65

 

= 2sin 

Relate two

 

M1

[Note: Diagram or otherwise defines v and w  perpendicular components may be be omitted]

A1

3.4

dimensional impacts to  problems involving two spheres of equal radius

Parallel to the line of centres forms equation using COL momentum formula using deduced values for cos α and cos β  cos β .

512  56   16   m   m w  m v  w  v  5 2 5 3 5 3    65  65   65 

3m  4 

Similarly forms equation using NEL formula: 256   56   16   4 2 0 . 5 2      w v      65  125     65  

M1 A1

3.4

A2

1.1b

M1

3.4

A1

3.2a

Thus solves the two simultaneous equations to find v and w. v  

96

w



325  and

2848 1625  oe (allow 3 decimal places or better)

Uses both components and total KE formula to calculate total KE lost,

3m 

 132  2  96  2   5m  2  126 2  2848  2  4    2    L          2  6 5 3 2 5 2 6 5 1625                   2

Or considers energy change in parallel components only

3m 

2  56 

2

 96   4      L =  2    65   325   Equates to given value

 L  

2 

5 m 

2  16

2

  2848   2         2   65   1625   

3  and solves:

m 

0.283  

2 

  

(8)

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4

 

Mark scheme

Further Mechanics 1 Unit Test 6: Elastic collisions in two dimensions

3b

Impulse acts parallel to the line of the centres (may be implied)

M1

3.1b

56   96  0.283   4  65     325 Thus for Q on  on P   P : Impulse

M1

1.1a

Thus magnitude of Impulse of Q on on P   P  =  = 0.892 Ns

A1

1.1b

 

5th Understand that impulse acts  perpendicular to the surface through the sphere's centre

(3) (11 marks) Notes

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5

 

Mark scheme

Further Mechanics 1 Unit Test 6: Elastic collisions in two dimensions

Q

4a

Scheme

Marks

Derives (or simply states) relationship between angles before Wall A:: (60) and after (  )  colliding with Wall A

AOs

6th M1

1.1a

tan     e  tan 60  e 3 Deduces for sphere to hit wall B wall  B::  0 „     45

0 „ tan     1 Thus:

 0„

e

1 3

*

Pearson Progression Step and Progress Descriptor

 

M1

2.2a

A1*

1.1b

Calculate the angle of  deflection in an oblique impact with a surface

(3) 4b

e 

Uses

1

    e 3  to find:     30 3  with tan   

Thus deduces sphere meets wall B wall  B at  at an angle:     15 tan  

Hence:

1 3

 

 

M1

1.1b

M1

1.1a

6th Calculate the angle of  deflection in an oblique impact with a surface

ta tan 1 15 5     5.104

 (5.1 or better)

A1

1.1b

(3)

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6

 

Mark scheme

Further Mechanics 1 Unit Test 6: Elastic collisions in two dimensions

4c

Uses parallel and perpendicular directions, and NEL formula, to find, for speed of sphere, v, after colliding with wall A wall  A:: v co sβ

 3 cco os6 0 

3 2   and

siin 6 60 0 v sin  β  3e s

v

Thus uses Pythagoras to find:

8th

3



3 cos15  w cos θ  and/or 3

3.3

M1

1.1b

2

= 3 oe

 

Solve a wide range of problems involving successive oblique impacts with smooth surfaces

Similarly establishes parallel and perpendicular relationships for  before and after the sphere collides collides with wall B wall  B,, where the sphere leaves with speed w, at an angle   , such that:

3

M1

M1

3.3

M1

1.1b

M1

1.1a

A1

1.1b

sin 15  w sin θ

2

Uses Pythagoras or    value from part 4bi to find w or w : w

2

2

1

2

2

 3 co cos 15  sin 15  w  2.82136  ( w  1.679  ) 3

Uses correct formula to find KE lost over both collisions, L collisions,  L:: 2 2 L  1  0.25  3  their w 2





 

.772  joules (accept 0.77 or better) Thus Overall KE lost, L  0.77

 

(6) (12 marks) Notes

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7

 

Mark scheme

Further Mechanics 1 Unit Test 6: Elastic collisions in two dimensions

Q

Scheme

Marks

5

Uses COL momentum formula along the line of centres using X  using  X   and Y  as  as the speeds of M  of  M  and N   and N  in  in that same direction after the collision respectively:

M1

2um cos 60 60  mX  mY  X  Y  u

A1

AOs

8th 3.3

 X 

M1A1

3.4

Solves these two equations simultaneously, and correctly for X  for  X : 1 1    X  u (1  e)[Y  u ( (1 1  e) not needed ] 2 2

A1

1.1b

Uses NEL result relating to angle    made by N by N relative to the   t a n  E tan 60  E  3        wall after the collision with it:

M1

3.4

After the collision with N  with  N , the perpendicular component of the   velocity of M  of M  is  is unchanged: 2u sin 60  u 3

M1

1.1a

M1

2.1

2u cos 60 60

= e  Y  X  ue

 

oe

Solve a wide range of problems involving oblique impacts between two spheres

Uses NEL formula   along line of centres Y

Pearson Progression Step and Progress Descriptor

The acute angle,   , from the line of centres made by M  by  M  after  after the collision, by trigonometry is thus: tan     

 perpendicularr component of velocity  perpendicula  parallel component of velocity



u

3

1 u (1  e) 2



2 3 1 e

 

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8

 

Mark scheme

Further Mechanics 1 Unit Test 6: Elastic collisions in two dimensions

Deduces from parallel line geometry if the paths are to be   α = 60  θ  parallel after the collisions then:

Thus:

tan α  tan  60 6 0    

tan 60  tan   1  tan 60 tan  



2 3

 

1   e

M1

2.2a

M1

2.1

M1

1.1b

A1*

2.2a

 Now substitutes tan      E  3  to give:

3   E  3 1  3 E



 E (1  e )  6 E

Hence:

2 3 1 e

   1   E   1  e   2(1  3E )

 2  (1  e)  E  

1 e 7e

*

 

(12 marks) Notes

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