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ZIMSEC A Level · 9164/4 · J2016

Mechanics Paper 4 June 2016

Questions
12
Total marks
24
Syllabus code
9164/4

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Questions
12
Pass mark
8
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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 7

[1 marks]Projectiles

The trajectory of a projectile is described by y=2x−0.01x2y = 2x - 0.01x^{2}, where x is the horizontal displacement and y is the vertical displacement from the point of projection.

Find the angle of projection, in degrees to the nearest degree.

Answer this when you sit the paper.

[2 marks]Projectiles

The trajectory of a projectile is described by y=2x−0.01x2y = 2x - 0.01x^{2}, where x is the horizontal displacement and y is the vertical displacement from the point of projection, and the angle of projection satisfies tan⁡θ=2\tan\theta = 2. Take g=9.81g = 9.81 ms−2^{-2}.

Find the initial velocity of projection, in ms−1^{-1}.

Answer this when you sit the paper.

[2 marks]Resultant of coplanar forces

Three coplanar forces act at a point Y. A force of 7 N is horizontal; a force of 6 N is inclined to it at 50∘50^{\circ} above the horizontal on the same side; and a force of 5 N acts at 30∘30^{\circ} below the horizontal, pointing away from Y in the opposite horizontal sense to the 7 N force.

Taking upwards as positive, find the vertical component of the resultant, in N.

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[2 marks]Resultant of coplanar forces

Three coplanar forces acting at a point Y have a resultant whose horizontal component is 6.53 N and whose vertical component is 2.096 N.

Find the magnitude of the resultant, in N.

Answer this when you sit the paper.

[1 marks]Resultant of coplanar forces

Three coplanar forces acting at a point Y have a resultant whose horizontal component is 6.53 N and whose vertical component is 2.096 N, both positive.

Find the angle, in degrees to the nearest degree, between the resultant and the horizontal.

Answer this when you sit the paper.

[2 marks]Kinematics

A particle starts from rest and accelerates at 2 ms−2^{-2} for 3 seconds. It then maintains the attained velocity for 4 seconds and then decelerates at 5 ms−2^{-2} for 2 seconds.

Find the velocity attained at the end of the first 3 seconds, in ms−1^{-1}.

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[2 marks]Kinematics

A particle starts from rest and accelerates at 2 ms−2^{-2} for 3 seconds, reaching 6 ms−1^{-1}. It maintains that velocity for 4 seconds, so it is still travelling at 6 ms−1^{-1} at t=7t = 7 s, and then decelerates at 5 ms−2^{-2} for 2 seconds.

Find the velocity of the particle at t=9t = 9 s, in ms−1^{-1}.

Answer this when you sit the paper.

[2 marks]Kinematics

A particle starts from rest and accelerates at 2 ms−2^{-2} for 3 seconds, reaching 6 ms−1^{-1}. It maintains 6 ms−1^{-1} for 4 seconds, then decelerates at 5 ms−2^{-2} for 2 seconds, so its velocity is −4-4 ms−1^{-1} at t=9t = 9 s and it is momentarily at rest at t=8.2t = 8.2 s.

Find the total distance travelled by the particle in the 9 seconds, in m.

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[2 marks]Kinematics

A particle travels 36.6 m forwards in the first 8.2 seconds of its motion, is momentarily at rest at t=8.2t = 8.2 s, and then travels 1.6 m backwards in the remaining 0.8 seconds.

Find the displacement of the particle at the end of the 9 seconds, in m.

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Section b, Question 8

[3 marks]Connected particles

Two identical small trays, each of mass 0.2 kg, are connected by a light inextensible string passing over a fixed smooth pulley, and they balance. A mass of 80 grammes is then placed on one of the trays, which begins to move downwards. Take g=9.81g = 9.81 ms−2^{-2}.

Find the acceleration of the trays, in ms−2^{-2}.

Answer this when you sit the paper.

[2 marks]Connected particles

Two trays, one of mass 0.2 kg and the other of mass 0.28 kg, hang from a light inextensible string over a fixed smooth pulley and accelerate at 1.635 ms−2^{-2}, the heavier one descending. Take g=9.81g = 9.81 ms−2^{-2}.

Find the tension in the string, in N.

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[3 marks]Connected particles

A mass of 80 grammes rests on a tray that is descending with an acceleration of 1.635 ms−2^{-2}. Take g=9.81g = 9.81 ms−2^{-2}.

Find the magnitude of the force exerted on the 80 gramme mass by the tray, in N.

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