Further Mechanics 1 Unit Test 4 Momentum And Impulse (part 2) Mark Scheme.docx

Mark scheme

Marks

AOs

Pearson Progression Step and Progress Descriptor

When t = 8, u1 = (8i + 2j) m s−1

M1

1.1a

4th

Also given: u2 = (4i + j) m s−1

M1

3.1a

M1

1.1b

A1

1.1a

Q

1a

Further Mechanics 1 Unit Test 4: Momentum and impulse (part 2)

Scheme

States that the combined particle has mass 2(3 × 10−6) = 6 × 10−6 and a single velocity, v. Use the vector form of the conservation of momentum:

Extend the definition of momentum and impulse to two dimensions

(3 × 10−6)(u1 + u2) = (6 × 10−6)v gives v = 6i + 1.5j Speed of new larger particle  36  2.25  6.18 m s−1 (3 s.f.)

(4) 1b

 1.5  6i  1.5 j  1  tan 1    6 

M1

2.2a

4th

A1

3.2a

3 2i  3 j   2  tan 1   2

M1

2.2a

Extend the definition of momentum and impulse to two dimensions

1  14.04 above i

 2  56.31 below i

A1

3.2a

Angle between two particles after collision = 14.04 + 56.31

A1

1.1b

= 70.4° (3 s.f.) (5) (9 marks) Notes

© Pearson Education Ltd 2018. Copying permitted for purchasing institution only. This material is not copyright free.

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Mark scheme

Further Mechanics 1 Unit Test 4: Momentum and impulse (part 2)

Marks

AOs

Pearson Progression Step and Progress Descriptor

Position of player with bat = 2i + j + 3(3i + 4j) = 11i + 13j

M1

1.1a

6th

Displacement player to skittle = 15i + 9j − (11i + 13j) = 4i − 4j

M1

1.1b

Velocity of ball after being struck by bat in order to hit the skittle is of the form ai − aj

M1

3.1b

This velocity has magnitude  5 2

M1

1.1b

M1

1.1a

A1

1.1b

Q

2a

Scheme

Magnitude of ai − aj =

Use the impulse/momentu m principle in vector form

2a 2  5 2

Hence a = 5 Use impulse−momentum principle in vector form. Impulse = 0.2((5i − 5j) − (3i + 4j)) Impulse = 0.4i − 1.8j

(6) 2b

Impulse given to ball from skittle = 0.2((−2i + 2j) − (5i − 5j))

M1

3.1a

= −1.4i + 1.4j Use Newton’s Third Law:

M1

3.2a

M1

1.1b

M1

1.1b

A1

1.1b

Impulse given to skittle from ball = 1.4i − 1.4j

6th Use the impulse/momentu m principle in vector form

0.8 × Force = 1.4i − 1.4j Force on skittle due to collision = 1.75i − 1.75j Magnitude of force on skittle = 2.475 N Maximum resistant force on skittle due to friction = μmg N = μ × 1 × 9.8 9.8μ = 2.475 μ = 0.253 (5) (11 marks) Notes

© Pearson Education Ltd 2018. Copying permitted for purchasing institution only. This material is not copyright free.

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