m2 collisions examwizard

City of London Academy 1

M2 COLLISIONS PAST QUESTIONS

1. A small ball A of mass 3m is moving with speed u in a straight line on a smooth horizontal

table. The ball collides directly with another small ball B of mass m moving with speed u

towards A along the same straight line. The coefficient of restitution between A and B is .2

1

The balls have the same radius and can be modelled as particles.

(a) Find

(i) the speed of A immediately after the collision,

(ii) the speed of B immediately after the collision. M

After the collision B hits a smooth vertical wall which is perpendicular to the direction of

motion of B. The coefficient of restitution between B and the wall is .5

2

(b) Find the speed of B immediately after hitting the wall. (2)

The first collision between A and B occurred at a distance 4a from the wall. The balls collide

again T seconds after the first collision.

(c) Show that T = .15

112

u

a

(6) (Total 15 marks)

2. Two particles, P, of mass 2m, and Q, of mass m, are moving along the same straight line on a

smooth horizontal plane. They are moving in opposite directions towards each other and collide.

Immediately before the collision the speed of P is 2u and the speed of Q is u. The coefficient of

restitution between the particles is e, where e < 1. Find, in terms of u and e,

(i) the speed of P immediately after the collision,

(ii) the speed of Q immediately after the collision. (Total 7 marks)

3. Particles A, B and C of masses 4m, 3m and m respectively, lie at rest in a straight line on a

smooth horizontal plane with B between A and C. Particles A and B are projected towards each

other with speeds u m s–1

and v m s–1

respectively, and collide directly.

As a result of the collision, A is brought to rest and B rebounds with speed kv m s–1

. The

coefficient of restitution between A and B is 4

3.


(a) Show that u = 3v. (6)

(b) Find the value of k. (2)

Immediately after the collision between A and B, particle C is projected with speed 2v m s–1

towards B so that B and C collide directly.

(c) Show that there is no further collision between A and B. (4)

(Total 12 marks)

4. A particle P of mass 3m is moving in a straight line with speed 2u on a smooth horizontal table.

It collides directly with another particle Q of mass 2m which is moving with speed u in the

opposite direction to P. The coefficient of restitution between P and Q is e.

(a) Show that the speed of Q immediately after the collision is 51 (9e + 4)u.

(5)

The speed of P immediately after the collision is 21 u.

(b) Show that e = 41 .

(4)

The collision between P and Q takes place at the point A. After the collision Q hits a smooth

fixed vertical wall which is at right-angles to the direction of motion of Q. The distance from A

to the wall is d.

(c) Show that P is a distance 5

3d from the wall at the instant when Q hits the wall.

(4)

Particle Q rebounds from the wall and moves so as to collide directly with particle P at the point

B. Given that the coefficient of restitution between Q and the wall is 5

1,

(d) find, in terms of d, the distance of the point B from the wall. (4)

(Total 17 marks)

5. A particle A of mass 4m is moving with speed 3u in a straight line on a smooth horizontal table.

The particle A collides directly with a particle B of mass 3m moving with speed 2u in the same

direction as A. The coefficient of restitution between A and B is e. Immediately after the

collision the speed of B is 4eu.


(a) Show that 4

3e .

(5)

(b) Find the total kinetic energy lost in the collision. (4)

(Total 9 marks)

6. Two small smooth spheres A and B have equal radii. The mass of A is 2m kg and the mass of B

is m kg. The spheres are moving on a smooth horizontal plane and they collide. Immediately

before the collision the velocity of A is (2i – 2j) m s–1

and the velocity of B is (–3i – j) m s–1

.

Immediately after the collision the velocity of A is (i – 3j) m s–1

. Find the speed of B

immediately after the collision. (Total 5 marks)

7.

2

A small smooth ball B, moving on a horizontal plane, collides with a fixed vertical wall.

Immediately before the collision the angle between the direction of motion of B and the wall is

2θ where 0o < θ < 45°. Immediately after the collision the angle between the direction of motion

of B and the wall is θ as shown in the diagram above. Given that the coefficient of restitution

between B and the wall is 8

3, find the value of tanθ.

(Total 8 marks)

8. A particle P of mass 2m is moving with speed 2u in a straight line on a smooth horizontal plane.

A particle Q of mass 3m is moving with speed u in the same direction as P. The particles collide

directly. The coefficient of restitution between P and Q is 2

1.

(a) Show that the speed of Q immediately after the collision is u3

8.

(5)


(b) Find the total kinetic energy lost in the collision. (5)

After the collision between P and Q, the particle Q collides directly with a particle R of mass m

which is at rest on the plane. The coefficient of restitution between Q and R is e.

(c) Calculate the range of values of e for which there will be a second collision between P

and Q. (7)

(Total 17 marks)

9. Two small spheres P and Q of equal radius have masses m and 5m respectively. They lie on a

smooth horizontal table. Sphere P is moving with speed u when it collides directly with sphere

Q which is at rest. The coefficient of restitution between the spheres is e, where e > 5

1.

(a) (i) Show that the speed of P immediately after the collision is .156

eu

(ii) Find an expression for the speed of Q immediately after the collision, giving your

answer in the form λu, where λ is in terms of e. (6)

Three small spheres A, B and C of equal radius lie at rest in a straight line on a smooth

horizontal table, with B between A and C. The spheres A and C each have mass 5m, and the

mass of B is m. Sphere B is projected towards C with speed u. The coefficient of restitution

between each pair of spheres is 5

4.

(b) Show that, after B and C have collided, there is a collision between B and A. (3)

(c) Determine whether, after B and A have collided, there is a further collision between B and

C. (4)

(Total 13 marks)

10. A particle P of mass m is moving in a straight line on a smooth horizontal table. Another

particle Q of mass km is at rest on the table. The particle P collides directly with Q. The

direction of motion of P is reversed by the collision. After the collision, the speed of P is v and

the speed of Q is 3v. The coefficient of restitution between P and Q is 2

1.

(a) Find, in terms of v only, the speed of P before the collision.


(3)

(b) Find the value of k. (3)

After being struck by P, the particle Q collides directly with a particle R of mass 11m which is

at rest on the table. After this second collision, Q and R have the same speed and are moving in

opposite directions. Show that

(c) the coefficient of restitution between Q and R is 4

3,

(4)

(d) there will be a further collision between P and Q. (2)

(Total 12 marks)

11. Two particles A and B move on a smooth horizontal table. The mass of A is m, and the mass of

B is 4m. Initially A is moving with speed u when it collides directly with B, which is at rest on

the table. As a result of the collision, the direction of motion of A is reversed. The coefficient of

restitution between the particles is e.

(a) Find expressions for the speed of A and the speed of B immediately after the collision. (7)

In the subsequent motion, B strikes a smooth vertical wall and rebounds. The wall is

perpendicular to the direction of motion of B. The coefficient of restitution between B and the

wall is 54 . Given that there is a second collision between A and B,

(b) show that 16

9

4

1 e .

(5)

Given that ,2

1e

(c) find the total kinetic energy lost in the first collision between A and B. (3)

(Total 15 marks)

12. A particle A of mass 2m is moving with speed 3u in a straight line on a smooth horizontal table.

The particle collides directly with a particle B of mass m moving with speed 2u in the opposite

direction to A. Immediately after the collision the speed of B is 38 u and the direction of motion


of B is reversed.

(a) Calculate the coefficient of restitution between A and B. (6)

(b) Show that the kinetic energy lost in the collision is 7mu2.

(3)

After the collision B strikes a fixed vertical wall that is perpendicular to the direction of motion

of B. The magnitude of the impulse of the wall on B is .3

14mu

(c) Calculate the coefficient of restitution between B and the wall. (4)

(Total 13 marks)

13. Two small spheres A and B have mass 3m and 2m respectively. They are moving towards each

other in opposite directions on a smooth horizontal plane, both with speed 2u, when they collide

directly. As a result of the collision, the direction of motion of B is reversed and its speed is

unchanged.

(a) Find the coefficient of restitution between the spheres. (7)

Subsequently, B collides directly with another small sphere C of mass 5m which is at rest. The

coefficient of restitution between B and C is 53 .

(b) Show that, after B collides with C, there will be no further collisions between the spheres. (7)

(Total 14 marks)

14. A particle P of mass 3m is moving with speed 2u in a straight line on a smooth horizontal table.

The particle P collides with a particle Q of mass 2m moving with speed u in the opposite

direction to P. The coefficient of restitution between P and Q is e.

(a) Show that the speed of Q after the collision is 51 u(9e + 4).

(5)

As a result of the collision, the direction of motion of P is reversed.

(b) Find the range of possible values of e. (5)


Given that the magnitude of the impulse of P on Q is 5

32 mu,

(c) find the value of e. (4)

(Total 14 marks)

15. [In this question i and j are perpendicular unit vectors in a horizontal plane.]

A ball has mass 0.2 kg. It is moving with velocity (30i) m s–1

when it is struck by a bat. The bat

exerts an impulse of (–4i + 4j) Ns on the ball.

Find

(a) the velocity of the ball immediately after the impact, (3)

(b) the angle through which the ball is deflected as a result of the impact, (2)

(c) the kinetic energy lost by the ball in the impact. (4)

(Total 9 marks)

16. Two small smooth spheres, P and Q, of equal radius, have masses 2m and 3m respectively. The

sphere P is moving with speed 5u on a smooth horizontal table when it collides directly with Q,

which is at rest on the table. The coefficient of restitution between P and Q is e.

(a) Show that the speed of Q immediately after the collision is 2(1 + e)u. (5)

After the collision, Q hits a smooth vertical wall which is at the edge of the table and

perpendicular to the direction of motion of Q. The coefficient of restitution between Q and the

wall is f, 0 < f 1.

(b) Show that, when e = 0.4, there is a second collision between P and Q. (3)

Given that e = 0.8 and there is a second collision between P and Q,

(c) find the range of possible values of f. (3)

(Total 11 marks)


17. A smooth sphere A of mass m is moving with speed u on a smooth horizontal table when it

collides directly with another smooth sphere B of mass 3m, which is at rest on the table. The

coefficient of restitution between A and B is e. The spheres have the same radius and are

modelled as particles.

(a) Show that the speed of B immediately after the collision is 41 (1 + e)u.

(5)

(b) Find the speed of A immediately after the collision. (2)

Immediately after the collision the total kinetic energy of the spheres is 61 mu

2.

(c) Find the value of e. (6)

(d) Hence show that A is at rest after the collision. (1)

(Total 14 marks)

18. A uniform sphere A of mass m is moving with speed u on a smooth horizontal table when it

collides directly with another uniform sphere B of mass 2m which is at rest on the table. The

spheres are of equal radius and the coefficient of restitution between them is e. The direction of

motion of A is unchanged by the collision.

(a) Find the speeds of A and B immediately after the collision. (7)

(b) Find the range of possible values of e. (2)

After being struck by A, the sphere B collides directly with another sphere C, of mass 4m and

of the same size as B. The sphere C is at rest on the table immediately before being struck by B.

The coefficient of restitution between B and C is also e.

(c) Show that, after B has struck C, there will be a further collision between A and B. (6)

(Total 15 marks)

19. A smooth sphere P of mass 2m is moving in a straight line with speed u on a smooth horizontal

table. Another smooth sphere Q of mass m is at rest on the table. The sphere P collides directly

with Q. The coefficient of restitution between P and Q is 31 . The spheres are modelled as

particles.

(a) Show that, immediately after the collision, the speeds of P and Q are 9

5 u and 9

8 u


respectively. (7)

After the collision, Q strikes a fixed vertical wall which is perpendicular to the direction of

motion of P and Q. The coefficient of restitution between Q and the wall is e. When P and Q

collide again, P is brought to rest.

(b) Find the value of e. (7)

(c) Explain why there must be a third collision between P and Q. (1)

(Total 15 marks)

20. A smooth sphere is moving with speed U in a straight line on a smooth horizontal plane. It

strikes a fixed smooth vertical wall at right angles. The coefficient of restitution between the

sphere and the wall is 21 .

Find the fraction of the kinetic energy of the sphere that is lost as a result of the impact. (Total 5 marks)

21. A smooth sphere S of mass m is moving with speed u on a smooth horizontal plane. The sphere

S collides with another smooth sphere T, of equal radius to S but of mass km, moving in the

same straight line and in the same direction with speed u, 0 < < 21 . The coefficient of

restitution between S and T is e.

Given that S is brought to rest by the impact,

(a) show that )1(

1

k

ke .

(6)

(b) Deduce that k > 1. (3)

(Total 9 marks)

1. (a) (i)

Con. of Mom: 3 3mu mu mv mw

2 3u v w (1) M1# A1

N.L.R: 12

u u w v M1# A1

u w v (2)


(1) – (2) 4u v DM1#

14

v u A1

(ii) In (2) 14

u w u

54

w u A17

(b) B to wall: N.L.R: 5 24 5u V M1

12

V u A1ft2

(c)

B to wall: 5 16

time 44 5

aa u

u B1ft

1 16 4

Dist. Travelled by 4 5 5

aA u a

u B1ft

In t secs, A travels1

4ut , B travels

1

2ut

Collide when speed of approach utut4

1

2

1 , distance to cover = M1$

aa5

44

45

34

4 16 4 64

5 3 15

a a a at

u u u

DM1$ A1

16 64 112

Total time 5 15 15

a a a

u u u

* A16

[15]

2.

21

2

2

vv

m

u

m

u

CLM: 4mu – mu = 2mv1 + mv2 M1 A1

i.e. 3u = 2v1 + v2


NIL: 3eu = –v1 + v2 M1 A1

v1 = u(1 – e) DM1 A1

v2 = u(1 + 2e) A1

[7]

3. (a)

Conservation of momentum: 4mu – 3mv = 3mkv M1A1

Impact law: vukv 4

3 M1A1

Eliminate k: 4mu – 3mv = 3m × vu 4

3 DM1

u = 3v (Answer given) A16

(b) 3,34

3 kvvkv M1,A12

(c) Impact law: (kv + 2v) e = vC – vB (5ve = vC – vB) B1

Conservation of momentum : 3× kv –1 × 2v =

3vB + vC (7v = 3vB + vc) B1

Eliminate vC : 05–74

ev

vB hence no

further collision with A. M1 A14 [12]

4. (a)

Correct use of NEL M1 *

y – x = e(2u + u) o.e. A1


CLM (→): 3m(2u) + 2m(–u) = 3m(x) + 2m(y) ( 4u = 3x + 2y) B1

Hence x = y – 3eu, 4u = 3(y – 3eu) + 2y, (u(9e + 4) = 5y) d * M1

Hence, speed of Q = ue )49(51 AG A1 cso5

(b) x = y – euueeu 3–)49(5

13 M1 *

Hence, speed )3–2(5

2)6–4(

5

1P e

uue o.e. A1

312,12–85)3–2(5

2

2

1 eeuuue

uux & solve for e d * M1

gives, e = 123

41e AG A1

Or

Using NEL correctly with given speeds of P and Q M1 *

3eu = uue21

51 –)49( A1

3eu = ,–21

54

59 uueu

21

54

5

9 –3 ee & solve for e d * M1

41

6015

103

56 eee . A14

(c) Time taken by Q from A to the wall =

u

d

y

d

5

4 M1

+

Distance moved by P in this time = )5

2

5

4

2(

2d

u

du

y

du

A1

Distance of P from wall = dddy

dd

53

52–;–

AG d

+M1;

A1 cso

or

Ratio speed P:speed Q = x:y = 5:24

5:

2

1)4

4

9(

5

1:

2

1 uuuu M1

+

So if Q moves a distance d, P will move a distance 52 d A1

Distance of P from wall = d – dd53;

5

2 AG cso d+M1;A14

(d) After collision with wall, speed uyQ u

41

4

5

51

51 their y B1ft


Time for P, ,–

21

5

3

u

xT

d

AB Time for Q, u

xTWB

41

from their y B1ft

Hence u

x

u

xTT

d

WBAB

41

21

53 –

M1

gives, dxxx,xxx dd51

5

3

5

3 32–4)–(2 5

3d A1 cao

or

After collision with wall, speed uyQ u

41

4

5

51

51 their y B1ft

speed u,xP21 speed P: new speed 1:2

4

1

2

1 u:uQ B1 ft

from their y

Distance of B from wall = 5

;5

3

3

1 dd their

12

1

M1; A14

2nd

or

After collision with wall, speed uyQ u

41

4

5

51

51 their y B1ft

Combined speed of P and uuuQ4

3

41

21

Time from wall to 2nd

collision = ud

ud

u

d

54

34

53

43

53

from their y B1ft

Distance of B from

wall = (their speed)x(their time) = du

du51;

5

4

4 M1; A14

[17]

5. (a) u 2u

4m 3m

x 4y eu=

3

LM 12mu + 6mu = 4mx + 12meu B1

NEL 4eu – x = eu M1A1

Eliminating x to obtain equation in e DM1

Leading to e = 4

3(*) cso A1 5


(b) x = 3eu or 4

9u or 4.5u – 3eu seen or implied in (b) B1

Loss in KE = 2

2

22 )3(32

1

4

94

2

1)2(3

2

1)3(4

2

1umumumum

M1A1ft

ft their x

222

8

3

8

52324 mumumu = 0.375mu

2 A1 4

[9]

6. 2m(2i – 2j) + m(–3i – j) = 2m(i – 3j) + mv M1A1

(i – 5j) = (2i – 6j) + v M1A1

(–i + j) = v A1

21)1( 22 v m s–1

cwo DM1A1 5

[5]

7. u cos 2 = v cos M1 A1

8

3u sin 2 = v sin M1 A1

3 tan 2 = 8 tan M1

tan8

tan1

tan62

M1

tan2 =

4

1 (tan 0)

tan = 2

1 M1 A1 8

[8]

8. (a) u u

2m 3m

x y

2

LM 4mu + 3mu = 2mx + 3my M1 A1

NEL uxy2

1 B1

Solving to y = *5

8u cso M1 A1 5

(b) x = u10

11 or equivalent B1


Energy loss

2

2

2

2

5

83

2

1

10

11)2(2

2

1uumuum M1 A(2,1,0)

2

20

9mu A1 5

(c) 85 u

3m m

s t

LM 5

24mu = 3ms + mt M1 A1

NEL eust5

8 B1

Solving to s = )3(3

2eu M1 A1

For a further collision )3(5

2

10

11euu M1

e > 4

1 ignore e ≤ 1 A1 7

[17]

9. u

m

V W

5m

(a) CLM: mv + 5mw = mu B1

NLI: w – v = eu B1

Solve v: v = 6

1(1 – 5e)u, so speed =

6

1(5e – 1)u M1*A1

(NB – answer given on paper)

Solve w: w = 6

1(1 + e)u M1*A16

* The M’s are dependent on having equations (not necessarily

correct) for CLM and NLI

(b) After B hits C, velocity of B = “v” = u)3

4.51(

6

1 = –½u M1A1

velocity < 0 change of direction B hits A A1CSO3


(c) velocity of C after = u10

3 B1

When B hits A, “u” = ½u, so velocity of B after = –½(–½u) = u4

1 B1

Travelling in the same direction but 10

3

4

1 no second collision M1A1cso4

B1 Conservation of momentum – signs consistent with their

diagram/between the two equations

B1 Impact equation

M1 Attempt to eliminate w

A1 correct expression for v. Q asks for speed so final answer must be

verified positive with reference to e > 1/5.

Answer given so watch out for fudges.

M1 Attempt to eliminate v

A1 correct expression for w

M1 Substitute for e in speed or velocity of P to obtain v in terms of u.

Alternatively, can obtain v in terms of w.

A1 (+/–)u/2

3

5wv

A1 CSO Justify direction (and correct conclusion)

B1 speed of C = value of w = 10

3)(

u (Must be referred to in (c)

to score the B1.)

B1 speed of B after second collision wu6

5)(or

4

1)(

M1 Comparing their speed of B after 2nd

collision with their speed of C

after first collision.

A1 CSO. Correct conclusion. [13]

10. (a)


m km

u

v 3v NEL 3v – (–V) = eu M1A1

u = 8v A1 3

(b) LM 8mv = –mv + 3kmv ft their u M1A1ft

(m × (u) = –mv + 3kmv)

k = 3 A13

(c)

3y

km 11m

LM 9mv = –3my + 11my ft their k M1A1ft

NEL 2y = e × 3v M1

y = 8

9v e =

4

3 * cso A14

(d) y = 8

9v > v further collision between P and Q M1A12

A1 is cso – watch out for incorrect statements re. velocity [12]

11. (a) u

v

m 4m

w

mu = 4mw – mv M1 A1

eu = w + v M1 A1

ue

vue

w )5

14(,)

5

1(

Indep M1 A1 A1 7

(b) ue

w )25

44(

B1 f.t.

Second collision w > v


5

14

25

44

ee M1

Also v > 0 e > 1 / 4 Hence result B1 5

(c) KE lost = ½ mu2 [½ 4m{(u / 5)(1 + e)}

2 + ½ m{(u/5)(4e 1)}

2] M1A1f.t.

2

10

3mu A1 cao 3

[15]

12. (a)

3u

x

2m

2u

8 /3u

m

LM 6mu – 2mu = 2mx + 38

mu M1 A1

ux

32

NEL 38

u – x = 5ue M1 A1

Solving to e = 52 M1 A1 6

(b) Initial K.E. = 21 × 2m(3u)

2 +

21 × m (2u)

2 = 11 mu

2

Final K.E. = 21 × 2m

2

32

u +

21 × m

2

38

u = 4mu

2 both M1

Change in K.E. = 7mu2 M1 Subtracting and simplifying M1 A1

to kmu2 A1cso 3

(c) m3

1438

vu mu M1 A1

(v = 2u)

e = 43

382 M1 A1 4

[13]


13. (a) 2u 2u

2u

3m 2m

v

CLM: 6mu – 4mu = 3mv + 4mu M1 A1

v = –3

2u A1

NLI: 2u – v = e.4u M1 A1

4eu = 3

8u e =

3

2. M1 A1 7

(b) 2u 0

y

2m 5m

x

5my + 2mx = 4mu M1 A1

y – x = 5

3.2u =

5

6u A1

Solve: x = –7

2u M1 A1

7

2u <

3

2u so B does not overtake A M1

So no more collisions A1cso 7 [14]

14.

3m 2m

2u u

x y

(a) LM 6mu – 2mu = 3mx + 2my M1 A1

NEL y – x = 3eu B1

Solving to y = 5

1u(9e + 4) (*) cso M1 A1 5

(b) Solving to x = 5

2u(2 – 3e) M1 A1

oe

x < 0 e > 3

2 M1 A1


3

2 < e 1 A1ft 5

ft their e for glb

(c) 2m[5

1u(9e + 4) + u] =

5

32mu M1 A1

Solving to e = 9

7 M1 A1 4

awrt 0.78 [14]

15. (a) I = mv – mu

–4i + 4j = 0.2v – 0.2 × 30i M1 A1

v = 10i + 20j (m s–1

) A1 3

(b) tan θ = 10

20 M1

θ = 63.4° A1 2

accept awrt 63° ot 1.1°

(c) Final K.E. = 2

1 × 0.2 × (10

2 + 20

2) (= 50) ft their v M1 A1ft

K.E. lost = 2

1 × 0.2 × 30

2 –

2

1 × 0.2 × (10

2 + 20

2) M1

= 40 (J) cao A1 4 [9]

16.

2m 3m

5u

x y

(a) LM 10mu = 2mx + 3my M1 A1

NEL y – x = 5eu B1

Solving to y = 2(1 + e)u (*) cso M1 A1 5

(b) x = 2u – 3eu finding x, with or without e = 0.4 M1


x = 0.8u A1

x > 0 P moves towards wall and Q rebounds from wall

second collision ft any positive x A1 ft 3

(c) x = – 0.4u B1

Speed of Q on rebound is 3.6fu

For second collision 3.6 fu > 0.4u M1

f > 9

1 ignore f 1 A1 3

[11]

17. (a) u 0 CLM: mu = mv1 + 3 mv2 B1

m 3m NIL: eu = v1 + v2 M1 A1

v1 v2 solving, dep. M1

v2 = 4

u(1 + e)* A1 5

(b) Solving for v1; )31(4

eu

M1 A1 2

(c) 2

1m

16

2u (1 3e)

2 +

2

13m

16

2u (1 + e)

2 =

6

1 mu

2 M1 A1 f.t. A1

e2 =

9

1 dep. M1 A1

e = 3

1 A1 6

(d) v1 = 4

u(1 3

3

1) = 0 at rest. A1 c.s.o. 1

[14]

18. (a)

m 2m BA e

u o

v v1 2

mu = mv1 + 2mv2 M1 A1

eu = v1 + v2 M1 A1

v1 = 3

u(1 – 2e); v2 =

3

u(1 + e) M1 A1 A1 7


(b) v1 > 0 3

u(1 – 2e) > 0 e <

21 M1 A1 c.s.o. 2

(c)

2m 4mB e

o

v v

v

3 4

2

2mv2 = 2mv3 + 4mv4 M1

ev2 = v3 + v4

v3 = 3

2v(1 – 2e) =

9

u(1 – 2e)(1 + e) M1 A1

Further collision if v1 > v3

i.e. if 3

u(1 – 2e) >

9

u(1 – 2e)(1 + e) M1

i.e. if 3 > 1 + e (as (1 – 2e) > 0)

i.e. if 2 > e M1

which is always true, so further collision occurs A1 cso 6 [15]

19. (a) L.M. 2u = 2x + y M1 A1

NEL y x = 3

1u M1 A1

Solving to x = 9

5u (*) M1 A1

y = 9

8u (*) A1 7

(b) () 9

8eu B1

L.M 9

10u

9

8eu = w M1 A1

NEL w = 3

1

euu

9

8

9

5 M1 A1

Solving to e = 32

25 M1 A1 7

accept 0.7812s

(c) Q still has velocity and will bounce back from wall

colliding with stationary P. B1 1 [15]

20. v = u2

1 B1


KE loss = m21 (u

2 – ( u

2

1 )2) M1

= 8

3 2mu A1

fraction of KE lost = 8

3 2mu

2

1mu

2 =

4

3 M1 A15

[5]

21. (a) u u mu + kmu = kmv

S m km T k

u(1 + k) = v M1 A1

0 v v = e(u – u) = eu(1 – ) M1 A1

k

u(1 + k) = eu(1 – ) M1

)1(

)1(

k

k = e(T) A16

(b) )1(

1

k

k 1 1 + k k(1 – ) M1

21

1

k A1

since 0 < < 2

1 , k > 1 (T) A13

[9]

1. There were also a significant number of fully correct answers to this question. Most candidates completed parts (a) and (b) well but part (c) proved to be rather more demanding.

For part (a) the majority of candidates understood and applied the conservation of linear momentum and the law of restitution

correctly. The equations were usually consistent despite the occasional lack of a clear diagram, and only a very small minority got the restitution equation the wrong way round.

However, subsequent errors in the speed of A and B were common – these arose from simple sign errors in the initial equations or

more commonly from minor processing errors. These basic errors can be costly – unexpected outcomes should be checked carefully

to avoid continuing to work with unrealistic situations. A few candidates surprisingly failed to substitute ½ in for e and then

struggled through with their answers for the rest of the question.

Full marks were usually scored for part (b) with only a minority of candidates with method errors in the use of the restitution equation. The question asked for the speed of B after the collision, so the final answer should have been positive, which was not

always the case.

There were a wide variety of approaches to part (c). It was pleasing to see that some candidates could produce the given expression fluently with their methods clearly laid out. The most successful approach involved finding the separation of the two particles as B

impacted with the wall and then to use the relative velocity to find the time taken to cover this separation. Some used the ratio of

distances travelled or set up an equation in T.


Many could find the time to B hitting the wall and the distance travelled by A in this time, but got no further, having run out of time

or having no idea how to proceed further.

A few solutions went off in entirely the wrong direction, either by thinking that another collision was needed (considering CLM and

NEL again) or by attempting to use methods which implied non-zero acceleration. A worrying handful did not use the correct

relationship between distance speed and time (e.g time = distance × speed was seen).

Expressions involving a and u were often badly written and these symbols would become interchanged during the rearrangement

and simplification of terms. For example, a fractional term such as u4

5 was written without due care so that the u migrated to the

bottom of the fraction during manipulation. Clear layout and a description of the symbols are vital in this kind of question to avoid

these careless errors and to help examiners navigate through a candidates’ work.

The given answer was helpful to some candidates who made a false start - realising their error they would often produce a better or

correct solution. However, candidates working with incorrect values from part (a) were often misled into altering work displaying

correct method in an attempt to derive the given answer. There were also some who tried to fudge incorrect processes to achieve the given answer.

2. This question was very well answered by the majority of candidates. The momentum and impact equations were often correct - the

most common error was lack of consistency in signs between their equations. A surprising number of candidates did not draw a

diagram which possibly made it more difficult to avoid these sign errors. Only a small number of candidates quoted the impact law the wrong way round.

A significant number of candidates had P going to the left after the collision, obtaining a velocity of u(e – 1). However, they almost

always failed to realise that this was a negative answer, ignoring the fact that the question had asked for speed. A

number of candidates, who started with two correct equations, went on to lose marks at the end due to algebraic errors in solving the

simultaneous equations.

3. Parts (a) and (b) were tackled with confidence by most candidates although a few were not sufficiently careful with signs and/or had long-winded algebraic manipulation to achieve the given result. CLM and the impact law were generally used correctly.

Part (c) proved to be more challenging and differentiated between the stronger candidates and those with confused concepts. No

mention of e in the question caused some to ignore it or, more commonly, to assume the previous value. There were various arguments used to justify their final statement but, encouragingly, the range for e seemed to be understood.

4. Candidates made a confident start to this question, but in parts (a) and (b), poor algebraic skills and the lack of a clear diagram with

the directions marked on it hampered weaker candidates’ attempts to set up correct and consistent (or even physically possible)

equations. The direction of P after impact was not given and those candidates who took its direction as reversed ran into problems when finding the value of e. Many realised that they had chosen the wrong direction and went on to answer part (b) correctly but

some did not give an adequate explanation for a change of sign for their velocity of P. Algebraic and sign errors were common, and

not helped by candidates’ determination to reach the given answers.

Parts (c) and (d) caused the most problems. They could be answered using a wide variety of methods, some more formal than others.

Many good solutions were seen but unclear reasoning and methods marred several attempts. Too many solutions were sloppy, with

u or d appearing and disappearing through the working. A few words describing what was being calculated or expressed at each stage would have helped the clarity of solutions greatly. Students need to be reminded yet again that all necessary steps need to be

shown when reaching a given answer. Too many simply stated the answer 5

3d without the explanation to support it.

5. (a) Very few candidates had a problem with the two equations required. The most common errors were still inconsistency in

signs and getting the ratio the wrong way round for the impact law. This is a very costly error leading to the loss of most

marks for this part of the question.

(b) Some candidates struggled with the algebra here. The correct values for the speeds of A and B immediately after the

collision were usually used, but, especially when candidates found the change in kinetic energy of the particles separately

and then combined their answers, they failed to subtract energies correctly. A significant number managed to make mass and velocity disappear from their final answer having successful worked through the algebra to find the kinetic energy lost.


6. Many candidates failed to recognise this as one of the most straightforward questions they are ever likely to meet at this level. Those candidates who did recognise the need for a simple application of conservation of momentum, using the vectors given, were few and

far between. It was more common for candidates to apply conservation of momentum in the i and j directions separately, and such

candidates generally did so successfully. Many candidates wanted this to be the more usual style of question, where they could work parallel and perpendicular to the line of centres of the two spheres. Candidates who tried to find the line of centres were generally

unsuccessful. Some assumed it was in either the i or the j direction and made no headway as a result. A large number of candidates

were content to regard the velocity as the speed and did not attempt to find the magnitude of their velocity.

7. This was another standard problem, which most candidates managed easily. The majority of candidates recognised the need to apply conservation of momentum parallel to the wall and Newton’s experimental law perpendicular to it. Those that did so often went on

to score full marks on this question.

There were many ways through the trigonometric manipulation including working with tan throughout, or solving for cos first and then finding tan. Errors involving the incorrect use of e were rare but sign errors in the use of Newton’s Experimental Law were

more common.

8. This question was generally well understood and answered. Most errors were caused by poor presentation leading to carelessness.

Candidates who kept all the velocities in the direction of the original velocities usually fared better than those who reversed one or

more velocity. The clearest solutions included clearly annotated diagrams which made the relative directions of motion very clear. In the weaker solutions it was sometimes difficult to work out the candidate’s thoughts about what happened in each collision -the

question did not give them names for the speeds after the initial collision and this gave rise to problems for some candidates who

often gave the same name to more than one variable. Candidates with an incorrect or inconsistent application of Newton’s Experimental Law lost a lot of time trying to obtain the given answer for the speed of Q after the first collision. In part (b) although

most candidates attempted to form a valid expression for the change in kinetic energy, the m and u2

were too often discarded along the way.

In part (c), and to a lesser extent in part (a), the tendency to want to solve simultaneous equations by substitution, rather than by

elimination, produced untidy and unwieldy expressions which often led to arithmetical errors; a shame when the original equations were correct. Most candidates interpreted the final part correctly, although too many wanted to substitute 11/10u rather than tackle

an inequality -it was clear that many candidates were not confident in setting up an inequality. Some problems did occur where

students assumed the reversal of the direction of motion of Q following the collision but failed to take account of this in setting up their inequality.

9. Very few candidates noticed the link between part (a) of this questions and parts (b) and (c).

This resulted in a considerable quantity of valid but unnecessary work. The marks allocated to the three parts of the question should

give candidates an indication that each of the later parts is not expected to involve as much work as the first part.

(a) There were many substantially correct answers to this part. Most candidates formed correct equations using restitution and

conservation of momentum. The difficulties started with the speed of P – many candidates whose answer was the negative

of the printed answer did not justify the change of sign using the information about the value of e, and those whose answer agreed did not appreciate the need to verify that their value for velocity was in fact positive.

Candidates who changed the sign of their answer for the speed of P often went on to substitute incorrectly to find the speed

of Q. There was also evidence of some confusion over the exact meaning of the question in part (aii), with several

candidates starting by substituting in place of e.

(b) Most candidates elected to start the question afresh rather than use the results from part (a). Examiners were presented with

confusing diagrams which were often contradicted by the working which followed. Alternatively there was no diagram and

we had to decide for ourselves which direction the candidate assumed sphere B would move in after the collision with C.

(c) By this stage in the working many candidates were working with an incorrect initial speed of B, and were then further

confused about the possible directions of motion of A and B after their collision. This often resulted in a page or more of

working to deduce a velocity for B after the collision with A, all for a potential score of one mark. Most candidates did demonstrate an understanding that they needed to compare the speeds of B and C to determine whether or not there would

be a further collision.


10. Many scored full marks in (a) and (b). It was pleasing to see that most used correct methods for momentum and impact law

equations but disappointing to see a number of sign and arithmetic errors. Parts (c) and (d) proved to be more challenging. It did not

help that some candidates confused themselves by giving every unknown speed the same name. Part (c) required the use of two

correct equations solved simultaneously and many successfully showed that e = ¾. In (d) marks were often lost through poor

explanation. For the final mark a clear statement backed up by a comparison of speeds was required.

11. In part (a) candidates generally understood the methods involved and were able to produce momentum and restitution equations. Inconsistencies between directions in diagrams and equations were common and many were unable to obtain a correct expression

for the speed of A after the collision. In the second part, the rebound again caused problems with signs but most were able to set up

the first inequality. The given double inequality was sometimes fudged or just stated without any attempt to justify e > ¼. In part (c), the correct method was usually adopted but accuracy errors were common.

12. The first part was very well answered with the odd error of incorrect signs. It was pleasing to note that only a few gave the

restitution equation the wrong way round. Most of the errors here occurred in parts (b) and (c). There appeared to be some confusion

over the fact that the KE of every particle before and after the impact was required and that these needed to be added and subtracted correctly in order to get a positive loss of KE. Some candidates missed out the KE of one of the particles and many subtracted to get

a negative KE. In part (c) the impulse equation was often incorrect, with an incorrect sign in the velocity term. This led to a value of

e >1. Some candidates realised that this was wrong and corrected their equation but others changed the final equation but failed to change the signs in their original impulse equation and were unable to score full marks.

13. The method in part (a) was generally well known and there were many good solutions. However, the second part proved to be more

challenging with only the better candidates able to set up and solve their equations correctly and even then not all were able to pick

up the final two marks - many only considered a further collision between B and C, which was impossible, instead of between A and B.

14. In part (a), nearly all candidates could obtain a pair of equations using conservation of linear momentum and Newton’s law of restitution. The printed answer usually helped those who had made sign errors to correct them. Part (b) proved more difficult. The

majority of candidates first found the velocity of P. A few, on recognising that the direction of P differed from the one they had

used in part (a), reversed the direction of P and started all over again rather than correctly interpreting the results they already had. In doing so they sometimes confused their two sets of working. Those who had found the velocity of P often gave the inequality the

wrong way round or produced fallacious work when solving the inequality. Another error, frequently seen, was to produce an

inequality from the incorrect reasoning that the speed of one particle had to be greater than the other. The fully correct range

23

1e was rarely given. One neat, although equivalent, method of solution seen was to say that the velocity of separation had

to be greater than the velocity of Q. This gives 15

3 9 4eu u e without the necessity of finding the velocity of P. Part (c)

is demanding in its algebraic requirements and the signs were difficult for many to sort out. However many completely correct

solutions were seen and the general standard of algebraic manipulation was good.

15. Part (a) was well done, although a minority changed the sign of the velocity from 130 msi to 130 ms i , thinking

that they were taking a “rebound” into account, not understanding that the direction was implied by the vector itself. In part (b) there

was some confusion between the angle of deflection and the angle the impulse made with i. Part (c) was quickly done by those who

knew how to find the kinetic energy associated with a velocity given in vector form but many could not do this.

16. Part (a) was well done but parts (b) and (c) proved testing. Many produced irrelevant equations for a second collision between the spheres and wasted much time not realising that the question was whether or not a second collision occurred and not what the results

of such a collision might be. The essential distinction between (b) and (c) is that in the former it is the direction of P after the first

collision which is important whereas in (c) it is the magnitude and this was not recognised by many. It was not unusual to see a


comparison of speeds in (b). In (c) there were many errors in manipulating the inequality and it was common to see the incorrect

inequality 3.6 0.4fu u leading, through an incorrect sequence of steps, to the “correct” 1

9f .

17. Part (a) was successfully completed by most of the candidates who attempted the question, although some thought that the two

particles had the same final speed. In part (b), most were able to obtain an expression for a velocity but the significance of the word ‘speed’ was not understood and there were perhaps only one or two correct answers from the entire candidature. In the third part

most candidates were unsuccessful in obtaining e=1/3 due to either forgetting that the mass of B was 3m or their inability to

accurately expand and collect terms from two squared brackets.

Others added the velocities of A and B together and squared the total and a significant number just “cancelled” the squares off the

three expressions in the energy equation reducing it to u = v1 + 3v2. Some may have misread the question and assumed that the

total KE before and after the collision was 1/6m u2 gaining an extra KE term. Some candidates realised from their answer to (b) that e should come out to be 1/3 and simply fiddled their working. Even some of those successful in (c) did not justify or show by

substitution that the velocity of A was zero, merely stating the fact without any reference even to e=1/3.

18. (a) The method here was generally well known and the momentum equation was usually correct although the NIL equation sometimes had sign errors. Manipulation of the equations was generally poor.

(b) This was rarely correct – a common error was to compare the two velocities. Many answers were unsupported and got no

credit.

(c) A significant number were able to write down attempts at the two equations but only a few were able to progress further as

the algebra became more complex. A tiny number realised that the second impact was essentially the same as the first and

were able to write down the final velocities without any further working. A full and convincing argument for this part was very rare.

19. The most notable characteristic about the majority of candidates’ responses to this question was the number of completely correct

solutions seen. The algebra in part (a) was quite straightforward but the same was definitely not true of part (b) and yet many could

negotiate the minimum of four variables needed in this question and obtain the correct exact answer 2532

e . The principles

needed to solve part (a) were well understood. When errors were made in part (b), these usually arose from not realising that there were two separate impacts to consider, one between Q and the wall, with an unknown e and a second impact between P and Q, in

which 13

e . Errors of sign were also seen, often resulting in a pair of incompatible equations for linear momentum and

Newton’s Experimental Law. In part (c), candidates were expected to explain that there would be a second impact between Q and

the wall, which would result in a further impact between Q and the stationary P. 20. No Report available for this question

21. No Report available for this question.

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