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ZIMSEC A Level · 9164/4 · N2009

Mechanics Paper 4 November 2009

Questions
10
Total marks
24
Syllabus code
9164/4

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Questions
10
Pass mark
6
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Answer every question in the printed order, get marked at the end, then see the answers.

The questions

Section b

Section b, Question 6

[3 marks]Motion on a rough inclined plane

A block of mass 3.5 kg is released from rest at point A on a plane inclined at angle α\alpha to the horizontal, where tan⁡α=34\tan\alpha = \dfrac{3}{4}. It slides down to point B at the base of the plane. The coefficient of friction between the block and the plane is 14\dfrac{1}{4}. Take g=9.81g = 9.81 ms−2^{-2}.

Calculate the acceleration of the block down the plane, in ms−2^{-2}.

Answer this when you sit the paper.

[2 marks]Motion on a rough inclined plane

A block of mass 3.5 kg is released from rest at point A on a plane inclined at angle α\alpha to the horizontal, where tan⁡α=34\tan\alpha = \dfrac{3}{4}. It slides down to point B at the base of the plane. The coefficient of friction between the block and the plane is 14\dfrac{1}{4}. Take g=9.81g = 9.81 ms−2^{-2}.

The block slides 5 m down the plane from A to B. Calculate the velocity of the block at B, in ms−1^{-1}.

Answer this when you sit the paper.

[2 marks]Motion on a rough inclined plane

A block of mass 3.5 kg is released from rest at point A on a plane inclined at angle α\alpha to the horizontal, where tan⁡α=34\tan\alpha = \dfrac{3}{4}. It slides down to point B at the base of the plane. The coefficient of friction between the block and the plane is 14\dfrac{1}{4}. Take g=9.81g = 9.81 ms−2^{-2}.

The block slides 5 m down the plane from A to B, reaching B at 6.264 ms−1^{-1}. Calculate the time it takes to slide from A to B, in seconds.

Answer this when you sit the paper.

Section b, Question 7

[2 marks]Friction on a horizontal wire

A ring of mass m kg accelerates at 1 ms−2^{-2} along a rough horizontal wire. The accelerating force of 6 N acts at 60∘^{\circ} above the wire, in the same vertical plane as the wire. The coefficient of friction between the ring and the wire is 14\dfrac{1}{4}.

Which expression gives the normal reaction R between the ring and the wire?

  1. AR=mg−33R = mg - 3\sqrt{3} newtons
  2. BR=mg−3R = mg - 3 newtons
  3. CR=mg+3R = mg + 3 newtons
  4. DR=mg+33R = mg + 3\sqrt{3} newtons
[2 marks]Friction on a horizontal wire

A ring of mass m kg accelerates along a rough horizontal wire. The normal reaction between the ring and the wire is R=mg−33R = mg - 3\sqrt{3} newtons and the coefficient of friction between the ring and the wire is 14\dfrac{1}{4}.

Which expression gives the frictional force acting on the ring?

  1. AF=4(mg−33)F = 4\left(mg - 3\sqrt{3}\right) newtons
  2. BF=14(mg+33)F = \dfrac{1}{4}\left(mg + 3\sqrt{3}\right) newtons
  3. CF=14(mg−33)F = \dfrac{1}{4}\left(mg - 3\sqrt{3}\right) newtons
  4. DF=14mgF = \dfrac{1}{4}mg newtons
[3 marks]Friction on a horizontal wire

A ring of mass m kg accelerates at 1 ms−2^{-2} along a rough horizontal wire. The accelerating force of 6 N acts at 60∘^{\circ} above the wire, in the same vertical plane as the wire. The coefficient of friction between the ring and the wire is 14\dfrac{1}{4}.

Find the value of m, giving your answer to 3 decimal places.

Answer this when you sit the paper.

[3 marks]Projectile motion

A particle is projected from a point O on the ground with speed V ms−1^{-1} at 60∘^{\circ} to the horizontal. It passes through the point A(3; 2)A(\sqrt{3};\ 2) on its way up, and reaches its maximum height above O at the point B.

Express V2V^{2} in terms of g.

  1. AV2=3gV^{2} = 3g
  2. BV2=12gV^{2} = 12g
  3. CV2=2gV^{2} = 2g
  4. DV2=6gV^{2} = 6g
[2 marks]Projectile motion

A particle is projected from a point O on the ground at 60∘^{\circ} to the horizontal with speed V ms−1^{-1}, and its speed satisfies V2=6gV^{2} = 6g. Take g=9.81g = 9.81 ms−2^{-2}.

Find the speed of projection V, in ms−1^{-1}.

Answer this when you sit the paper.

[2 marks]Projectile motion

A particle is projected from a point O on the ground at 60∘^{\circ} to the horizontal with speed V ms−1^{-1}, where V2=6gV^{2} = 6g. B is the point at which the particle reaches its maximum height above O.

Find the maximum height reached, that is the y coordinate of B, in metres.

Answer this when you sit the paper.

[3 marks]Projectile motion

A particle is projected from O at 60∘^{\circ} to the horizontal with V2=6gV^{2} = 6g. It passes through A(3; 2)A(\sqrt{3};\ 2) and reaches its maximum height at B(332; 2.25)B\left(\dfrac{3\sqrt{3}}{2};\ 2.25\right).

Find the angle, to the nearest degree, that the straight line AB makes with the horizontal.

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