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ZIMSEC A Level · 9164/4 · J2008

Mechanics Paper 4 June 2008

Questions
8
Total marks
12
Syllabus code
9164/4

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Questions
8
Pass mark
5
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Answer every question in the printed order, get marked at the end, then see the answers.

The questions

Question 1201

[1 marks]statics - friction
A man pulls a crate of weight 20 N along a rough horizontal floor using a string inclined at 60° to the horizontal. The crate is in limiting equilibrium when he pulls with a force of magnitude 4 N. Find the exact normal reaction between the crate and the floor.
  1. A(20+23)(20+2\sqrt3) N
  2. B(20−43)(20-4\sqrt3) N
  3. C(20−23)(20-2\sqrt3) N
  4. D(20−22)(20-2\sqrt2) N

Question 1202

[1 marks]statics - friction
Continuing the crate scenario (weight 20 N, pulling force 4 N at 60° to the horizontal, normal reaction R=20−23R=20-2\sqrt3 N, limiting equilibrium), find the exact coefficient of friction between the crate and the floor.
  1. A40−4340-4\sqrt3
  2. B10+397\dfrac{10+\sqrt3}{97}
  3. C3+10397\dfrac{3+10\sqrt3}{97}
  4. D110\dfrac{1}{10}

Question 1301

[1 marks]kinematics
A car travels at 40 ms−1^{-1} and decelerates uniformly to 25 ms−1^{-1} over 30 seconds, then continues at the constant speed of 25 ms−1^{-1} for a further distance of 500 m. Find the total distance travelled by the car.
  1. A14751475 m
  2. B12501250 m
  3. C975975 m
  4. D17001700 m

Question 1302

[1 marks]kinematics
For the same journey (uniform deceleration from 40 to 25 ms−1^{-1} over the first 30 s, then constant 25 ms−1^{-1} for the next 500 m), which statement correctly describes the shape of the displacement-time (t,x)(t,x) graph?
  1. AA curve that is concave down (decreasing gradient) for 0≤t≤300\le t\le30, followed by a straight line of constant gradient 25 for t>30t>30.
  2. BA curve that is concave up (increasing gradient) for 0≤t≤300\le t\le30, followed by a straight line for t>30t>30.
  3. CA straight line throughout, since the two legs of the journey combine into a single linear relationship.
  4. DA straight line for 0≤t≤300\le t\le30, followed by a curve that is concave up for t>30t>30.

Question 1401

[1 marks]projectile motion
A stone is thrown from the top of a vertical wall 3 m high with velocity VV ms−1^{-1} at an angle of depression θ°\theta° below the horizontal. It lands 0.5 s later at a point 1.6 m horizontally from the foot of the wall. Taking g=9.8g=9.8 ms−2^{-2} and neglecting air resistance, find θ\theta.
  1. Aθ≈42.0°\theta\approx42.0°
  2. Bθ≈7.1°\theta\approx7.1°
  3. Cθ≈48.0°\theta\approx48.0°
  4. Dθ≈61.9°\theta\approx61.9°

Question 1402

[1 marks]projectile motion
For the same stone, thrown at an angle of depression θ\theta from the top of a 3 m wall and landing 0.5 s later 1.6 m from the foot of the wall, with horizontal and vertical velocity components Vcos⁡θ=3.2V\cos\theta=3.2 ms−1^{-1} and Vsin⁡θ=3.55V\sin\theta=3.55 ms−1^{-1}, find VV.
  1. AV≈3.2V\approx3.2 ms−1^{-1}
  2. BV≈4.78V\approx4.78 ms−1^{-1}
  3. CV≈6.75V\approx6.75 ms−1^{-1}
  4. DV≈3.55V\approx3.55 ms−1^{-1}

Question 1501

[1 marks]connected particles - mechanics
A toy car P (mass 4 kg) rests on a rough horizontal plane, with coefficient of friction 12\frac12 between P and the plane, and is connected by a light inextensible string over a smooth pulley X to toy car Q (mass 8 kg) held on a smooth plane inclined at 30° to the horizontal. When Q is released, find the acceleration of the system and the tension in the string, in terms of gg.
  1. Aa=g6a=\dfrac{g}{6}, T=8g3T=\dfrac{8g}{3}
  2. Ba=g2a=\dfrac{g}{2}, T=4gT=4g
  3. Ca=0a=0, T=4gT=4g
  4. Da=g6a=\dfrac{g}{6}, T=2g3T=\dfrac{2g}{3}

Question 1502

[1 marks]connected particles - mechanics
For the same system (toy cars P and Q connected over smooth pulley X, tension T=8g3T=\dfrac{8g}{3} N in the string on both sides of the pulley, one side horizontal to P and the other along the 30° incline to Q), find the magnitude of the force exerted by X on the string, giving the answer in the form kgcos⁡αkg\cos\alpha.
  1. A16g3cos⁡30°\dfrac{16g}{3}\cos30°
  2. B8g3cos⁡75°\dfrac{8g}{3}\cos75°
  3. C16g3cos⁡15°\dfrac{16g}{3}\cos15°
  4. D16g3cos⁡75°\dfrac{16g}{3}\cos75°

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