Further Mechanics 1 Unit Test 3 Elastic Collisions in One Dimension Mark Scheme

 

Mark scheme

Further Mechanics 1 Unit Test 3: Elastic collisions in one dimension

Q

1a

Scheme

Uses conservation of momentum to form equation

(1. 1.5 5)(0.4 )(0.4))  (4. (4.5)(0 5)(0.1 .1))  1.5v1  4. 4.5 5v2   or 1.5v1  4.5v2





v1

0.4 0.4  ( 0.1 0.1)

 



1   or

M1

3.1b

3rd Understand the

M1

1.1a

A1

1.1b

concept of the elasticity and coefficient of  restitution

v2  v1  0.5

Solves simultaneously to find

v1    0.35   and v2    0.15  

  1b

AO

0.15

Uses e =1 with Newton’s law of restitution to form equation:

v2

Marks

Pearson Progression Step and Progress Descriptor

(3)

Forms equation in terms of e using Newton’s law of restitution:

v2  v1 0. 0.4 4  ( 0.1 0.1)



e   or

M1

3.4

4th Use Newton's experimental law

v2  v1  0.5e

Uses new equation and conservation of momentum equation to find that:

M1

3.4

M1

1.2

A1

1.1b

of restitution direct impactsfor  of  elastic spheres

0.12 125 5e  0. 0.0 025   v1   0.025 0.025  0.375 0.375e   and v2   0. Deduces from question that

v1  v2  

Forms 0.375e  0.025  0.125e  0.025,  solves to find

e  

1 5

(4)

1c

Shows clearly that because e …0  then 0.125e v2    0  for any permissible value of e



0.025



0  thus

v Thus, by considering 1  critical value of 0.025  0.375e  0   defines range of values for motion in opposite directions as 1   e „ 1 15

B1

2.1

B1

2.1

3rd Understand the range of values that the coefficient of  restitution can take

(2) (9 marks) Notes

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Mark scheme

Further Mechanics 1 Unit Test 3: Elastic collisions in one dimension

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2

 

Mark scheme

Further Mechanics 1 Unit Test 3: Elastic collisions in one dimension

Q

2a

Scheme

Marks

AOs

Speed of ball as it hits the horizontal plane using suvat:

v2



02



5th

 

2 gh0   or v  2 gh0

Uses Newton’s law of restitution to deduce speed of ball at instant after first collision:

u1  e 2 gh0



M1

3.3

Solve problems of 

A1

1.1b

 balls bouncing off  horizontal elastic  planes

M1

1.1a

A1

1.1b

 

Uses suvat to find maximum height after first collision, 0  e 2 gh0

Pearson Progression Step and Progress Descriptor

 2





h1 :

2 gh1

Rearranges to show h1



 

2

e h0

(4) 2b

Deduces that part a answer is part of an inductive sequence:

M1

2.2a

hn1   e 2 hn   Thus, derives formula in terms of e  and h0 : hn



(e 2 ) n h0  

A1

2.2a

6th Solve problems involving successive collisions including collisions with walls

(2) 2c

4

0.5ee h0 Interprets question to form equation: 0.5 Solves: e

 

0.917



e12 h0

 

 

States that the collisions are highly elastic, i.e. e  close to 1. 

M1

3.1b

A1

1.1b

B1

3.2a

6th Solve problems involving successive collisions including collisions with walls

(3) (9 marks) Notes

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3

 

Mark scheme

Further Mechanics 1 Unit Test 3: Elastic collisions in one dimension

Q

3a

Scheme

States velocity of 2kg sphere after first impact with wall



6e  

Uses KE formula to form an equation for energy lost:

1  E1    ( 2)[62 2



2

Marks

AOs

Pearson Progression Step and Progress Descriptor

M1

1.2

5th

A1

1.1a

Calculate the change as in akinetic energy result of a collision

2

(6e) ]  36(1  e ) (2)

3b

Uses conservation of momentum to form an equation eg.

3.3

5th

M1

3.4

Calculate the change in kinetic energy as a result of a collision

A1

1.1b

M1

1.1a

M1

1.1a

A1

1.1b

M1

12e  2v1  4v2

 

Forms an equation using Newton’s law of restitution eg.

6e

2



v2



v1

Solves for expressions for

v1  and v2 :

v1   2e(1  2e)  oe

v2   2 e (e  1)   oe

and

Uses correct KE loss equation:

 E2



1 2

( 2)[(6e) 2

  36 e Expands:  E2  Thus  E2



2



1 ( 2e (1  2e)) 2 ]  (4)[0 2  (2e( e  1))]2 2



4e 2 (1  4e  4e 2 )  8e 2 (e 2



2e  1)

2 4e 2 (1  e 2 )  *  (must show intermediate step)

 

(6) 3c

Forms equation using their  E1 2

2



E 2    16

3.1b

5th

M1

1.1a

Calculate the change in kinetic energy as a result of a collision

A1

1.1b

M1

2

36(1  e )  24e (1  e )  16 4 Forms quartic: 6e



2 Forms quadratic 6k

3e 2 



50

 

3k   5  0  and square roots positive

solution, or otherwise, to find: e   0.835 (3) (11 marks) Notes:

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Mark scheme

Further Mechanics 1 Unit Test 3: Elastic collisions in one dimension

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5

 

Mark scheme

Further Mechanics 1 Unit Test 3: Elastic collisions in one dimension

Q

4a

Scheme

Marks

Forms two equations using conservation of momnetum momnetum and  Newton’s law of restitution for example:

w B



wA



ue   and 4w B



w B



1 3

u (1  e )

M1

3.4

A1

1.1b

M1

3.4

M1

3.4

M1

1.1b

A1

1.1b

 

w

Uses  B  above as initial velocity with Newton’s law of restitution to form one equation for the collision between B between  B and  and v B

6th

2wA  2u

Thus, velocity of B of B after  after collision with A with A,,

C , using

AOs

Pearson Progression Step and Progress Descriptor

Solve problems involving successive collisions including collisions with walls

 for the velocity of B of B after  after this collision for example: 1 3

u e(1   e )  vC



vB

Forms second equation using conservation of momentum eg.

4 u (1  e  )  6vC 3



4v B

Solves simultaneously to show:

vC  

2 15

u (1  e) 2

(Please see notes section also for this mark)

Uses substitution to derive:

v B



1 15

u ( 2  e  3e 2 )   AG

(6) 4b

Deduces that for A for A and  and B  B to  to collide again whilst travelling in the w  vB same direction as C  then  then  A

Thus:

1 1 (1  2e )  ( 2  e  3e 2 ) 3 15

M1

2.2a

M1

1.1a

M1

1.1a

A1

1.1b

A1

2.2a

Rearranges to form quadratic inequality:

5  10e  2  e  3e 2



3e 2



9e  3  0

 

Solves quadratic inequality: e   0.382   or e   2.618 States possible range for e:

0  e  0.382

   

5th Solve problems involving successive collisions of pairs of spheres in one dimension

(5)

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Mark scheme

Further Mechanics 1 Unit Test 3: Elastic collisions in one dimension

(11 marks) Notes v

If the student derives  B  correctly in 4a without finding working must be very clear.

vC 

 award M1A1 for correct

v B

 expression but must the

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7

 

Mark scheme

Further Mechanics 1 Unit Test 3: Elastic collisions in one dimension

Q

5a

Scheme

Marks

AOs

Pearson Progression Step and Progress Descriptor

7th

Resolves to find forces in direction of motion: 2 Force due to friction  ke g cos(10)  and Force down slope due to gravity   g  sin(10)

M1

3.3

involvingin collisions unfamiliar  contexts

Energy lost in journey down slope = Work done against friction:

  2 cos(10) Frictional force   distance = ke g  

Solve problems

A1

3.4

M1

3.4

M1

1.1b

M1

3.4

M1

1.1b

A1

1.1b

Equates energy or, using suvat, deduces the speed before the collision with buffer,

w

2 g sin ((1 10)  ke 2 g co  s (1 (10)  

Thus, deduces the speed after collision:  v  e

2 g sin (1 (10)  ke 2 g  cos (1 (10)

 

 

Uses KE formula with v  and collision with buffer: 1 2

(1)[ w2

 to find energy lost in the



1 v 2 ]  [(2 g sin (10)  2ke2 g co  s (10)) 2



e2 (2 g sin (10)  2ke 2 g co   s (1 (10))]

Simplifies to show: 1 (1)[ w2  v 2 ]  (1  e2 )( g sin (10)  ke 2 g co  s (10))   2 So total energy lost from work done against friction and collision:

 L  (1  e 2 )( g sin (1 (10)  ke 2 g cos (1 (10))  ke 2 g  cos (1 (10)   2

4

So  L  (1  e ) g sin (10)  e kg cos (10)   AG (7)

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Mark scheme 5b

Further Mechanics 1 Unit Test 3: Elastic collisions in one dimension

7th

Interprets question to form equation:

1    g sin ((1 10)  (1  e 2 ) g sin ((1 10)  e 4 kg  cos (1 (10) 2   4 ( c o t ( 1 0 ) ) k e Rearranges to form quartic:



e

2



0. 5  0

 

M1

3.1b

M1

1.1a

A1

2.2a

Solve problems involving collisions in unfamiliar  contexts



Uses discriminant: 1 2k  c o ot  t (1 (10) …0  to state:

k    Maximum possible value of

1 2

tan (1 (10)  0.088  to 3dp. (3) (10 marks) Notes

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