Mechanics: Projectiles and Applications Depth: Fluency and Exam Drill - Worksheets, Questions and Revision

19 original exam-style questions - 4 pages of questions with a full mark scheme - free printable PDF.

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M8D Mechanics: Projectiles and Applications Depth: Fluency and Exam Drill

EDEXCEL 9MA0 · Calculator allowed · about 50 minutes

Total Marks

Name: _______________________________ Date: ____ / ____ / ______

Answer ALL questions. Show all your working.

A particle is projected from ground level with speed 15 m/s at an angle of 40 degrees above the horizontal. Find the horizontal component of its initial velocity, giving your answer to 3 significant figures.

(Total for Question 1 is 1 mark)

A particle is projected with speed 15 m/s at an angle of 40 degrees above the horizontal. Find the vertical component of its initial velocity, giving your answer to 3 significant figures.

(Total for Question 2 is 1 mark)

The vertical component of a particle's initial velocity, as it is projected under gravity, is 19.6 m/s. Taking g = 9.8 m/s², find the time taken for the particle to reach its greatest height.

(Total for Question 3 is 1 mark)

State the acceleration acting on a particle at every instant after it has been projected and is moving freely under gravity, giving your answer in the form ...j, using g = 9.8 m/s² and taking j as the vertically upward unit vector.

(Total for Question 4 is 1 mark)

A particle is projected from ground level with speed 12 m/s at an angle of 30 degrees above the horizontal, and moves freely under gravity. Take g = 9.8 m/s². Find the greatest height reached by the particle, giving your answer to 3 significant figures.

(Total for Question 5 is 2 marks)

A particle is projected from ground level with speed 9.8 m/s at an angle of 30 degrees above the horizontal, and lands on the same horizontal ground. Take g = 9.8 m/s². You may quote the result T = 2Usin(α)/g without proof. Find the time of flight, giving your answer exactly.

(Total for Question 6 is 1 mark)

A particle is projected from ground level with speed 14 m/s at an angle of 45 degrees above the horizontal, and lands on the same horizontal ground. Take g = 9.8 m/s². You may quote the result R = U² sin(2alpha)/g without proof. Find the horizontal range, giving your answer exactly.

(Total for Question 7 is 1 mark)

A particle is projected from ground level at an angle of 50 degrees above the horizontal, and lands on the same horizontal ground at a horizontal distance of 32 m from the point of projection. Take g = 9.8 m/s². You may quote the result R = U² sin(2alpha)/g without proof. Find the initial speed U, giving your answer to 3 significant figures.

(Total for Question 8 is 2 marks)

A stone is thrown horizontally with speed 8 m/s from a bridge 19.6 m above a river, and moves freely under gravity until it lands in the river. Take g = 9.8 m/s². Find the horizontal distance from the bridge at which the stone lands, giving your answer exactly.

(Total for Question 9 is 2 marks)

A particle P is projected from the origin O with velocity (5i + 9.8j) m/s, and moves freely under gravity, taking g = 9.8 m/s² acting in the direction -j. Find the velocity of P, in terms of i and j, when t = 1.5 seconds.

(Total for Question 10 is 2 marks)

A particle is projected from the origin with velocity (6i + 12j) m/s, and moves freely under gravity, taking g = 9.8 m/s² acting in the direction -j. Find the speed of the particle when t = 1 second, giving your answer to 3 significant figures.

(Total for Question 11 is 2 marks)

A particle is projected from a point on horizontal ground with speed U at an angle of 20 degrees above the horizontal, and moves freely under gravity, landing on the same horizontal ground. The greatest height reached by the particle is 4.9 m. Take g = 9.8 m/s². Find the value of U, giving your answer to 3 significant figures.

(Total for Question 12 is 2 marks)

A particle is projected from ground level with speed 24.5 m/s at an angle α above the horizontal, and moves freely under gravity, landing on the same horizontal ground. Take g = 9.8 m/s². Given that the range is a maximum for this speed, find the value of the maximum range, giving your answer exactly.

(Total for Question 13 is 2 marks)

A particle is projected from ground level with speed 24 m/s at an angle of 55 degrees above the horizontal, and moves freely under gravity, landing on the same horizontal ground. Take g = 9.8 m/s². Find the maximum height reached and the total time of flight, giving each answer to 3 significant figures.

(Total for Question 14 is 3 marks)

A ball is thrown horizontally with speed 14 m/s from the top of a cliff 44.1 m above the sea, which may be modelled as a horizontal plane. The ball moves freely under gravity. Take g = 9.8 m/s². Find the speed of the ball as it hits the sea, giving your answer to 3 significant figures.

(Total for Question 15 is 3 marks)

A particle P is projected from the origin O with velocity (7i + 24.5j) m/s, and moves freely under gravity, taking g = 9.8 m/s² acting in the direction -j. At time t seconds after projection, the position vector of P relative to O is r metres. Find the position vector of P, in the form ai + bj, at the instant it is moving parallel to the vector (2i - 3j).

(Total for Question 16 is 3 marks)

A particle is projected from a point O with speed 18 m/s at an angle of 40 degrees above the horizontal. The point O lies at the top of a straight slope which descends at a constant angle of 20 degrees below the horizontal, in the same vertical plane as the particle's motion, so that the particle lands at a point P on the slope. Horizontal and vertically-upward displacements from O are denoted x and y respectively (both in metres). Take g = 9.8 m/s². Given that the slope satisfies the equation y = -x tan(20), find the time of flight from O to P, giving your answer to 3 significant figures.

Figure (to be drawn): Diagram: point O at the top of a downward slope inclined at 20 degrees to the horizontal; a particle is projected from O at 40 degrees above the horizontal, in the same vertical plane as the slope, and lands on the slope at P.

(Total for Question 17 is 3 marks)

A crate of mass 4 kg is released from rest at the top of a rough plane inclined at 20 degrees to the horizontal. The coefficient of friction between the crate and the plane is 0.3. The crate slides a distance of 3.5 m down the plane before reaching the bottom. Take g = 9.8 m/s². Find the acceleration of the crate down the plane, and its speed at the bottom of the plane, giving each answer to 3 significant figures.

(Total for Question 18 is 4 marks)

Block P, of mass 2 kg, lies on a rough horizontal table. A light inextensible string connects P, over a smooth light pulley fixed at the edge of the table, to block Q, of mass 5 kg, which hangs freely. The coefficient of friction between P and the table is 0.25. The system is released from rest with the string taut, and Q descends, pulling P towards the pulley. Take g = 9.8 m/s². Find the acceleration of the system, and the tension in the string, while both blocks are moving.

Figure (to be drawn): Diagram: block P on a rough horizontal table, connected by a string over a smooth pulley at the table edge to block Q, which hangs freely and descends when the system is released.

(Total for Question 19 is 4 marks)

Mark scheme · M8D Mechanics: Projectiles and Applications Depth: Fluency and Exam Drill

Question 1

B1 11.5 cao
Answer: 11.5 m/s

Question 2

B1 9.64 cao
Answer: 9.64 m/s

Question 3

B1 2 cao
Answer: t = 2 s

Question 4

B1 -9.8j cao
Answer: -9.8j m/s²

Question 5

M1 substitutes U = 12, α = 30 into H = U² sin²(α)/(2g) (oe)
A1 H = awrt 1.84 (m), cao
Answer: H = awrt 1.84 m

Question 6

B1 1 cao
Answer: T = 1 s (exact)

Question 7

B1 20 cao
Answer: R = 20 m (exact)

Question 8

M1 rearranges R = U² sin(2alpha)/g to U² = Rg/sin(2alpha) and substitutes R = 32, α = 50 (oe)
A1 U = awrt 17.8 (m/s), cao
Answer: U = awrt 17.8 m/s

Question 9

M1 uses s = 4.9t² = 19.6 to find the time of fall, t = 2 (s)
A1 16 (m), cao
Answer: 16 m (exact)

Question 10

M1 uses v = u + at componentwise, with a = -9.8j
A1 5i - 4.9j (m/s), cao
Answer: v = 5i - 4.9j (m/s)

Question 11

M1 finds v(1) = 6i + (12 - 9.8)j = 6i + 2.2j, using v = u + at componentwise
A1 awrt 6.39 (m/s), cao
Answer: awrt 6.39 m/s

Question 12

M1 rearranges H = U² sin²(α)/(2g) to U = √2gH/sin(α) and substitutes H = 4.9, α = 20 (oe)
A1 U = awrt 28.7 (m/s), cao
Answer: U = awrt 28.7 m/s

Question 13

M1 recognises the range is greatest when α = 45 degrees, giving R_max = U²/g
A1 61.25 (m), exact, cao
Answer: R_max = 61.25 m (exact)

Question 14

M1 finds the initial vertical component of velocity, u_y = 24 sin(55) (oe)
A1 H = awrt 19.7 (m), using H = u_y²/(2g)
A1 T = awrt 4.01 (s), using T = 2u_y/g
Answer: H = awrt 19.7 m; T = awrt 4.01 s

Question 15

M1 uses s = 4.9t² = 44.1 to find the time of fall, t = 3 (s)
M1 finds the vertical speed at impact v_y = gt = 9.8 x 3 = 29.4, and combines with the horizontal speed using Pythagoras, speed = √14² + 29.4²
A1 awrt 32.6 (m/s), cao
Answer: 32.6 m/s (3sf)

Question 16

M1 uses v = u + at to find v(t) = 7i + (24.5 - 9.8t)j
M1 sets up the condition for v to be parallel to 2i - 3j, e.g. 7(-3) - (24.5 - 9.8t)(2) = 0, and solves to find t = 25/7 (or awrt 3.57)
A1 substitutes their t into r(t) = 7ti + (24.5t - 4.9t²)j to find r = 25i + 25j (m), cao (exact)
Answer: r = 25i + 25j (m), at t = 25/7 s (awrt 3.57 s)

Question 17

M1 substitutes x = 18cos(40) t and y = 18sin(40) t - 4.9t² into the slope equation y = -x tan(20)
M1 collects terms, factors out t (noting t is not 0 at P) and rearranges to make t the subject, e.g. t = (18sin40 + 18cos40 tan20)/4.9
A1 t = awrt 3.39 (s), cao
Answer: t = awrt 3.39 s

Question 18

M1 resolves perpendicular to the plane to find R = 4g cos(20), and parallel to the plane to find 4g sin(20) - 0.3(4g cos(20)) = 4a
A1 a = g(sin20 - 0.3cos20) = awrt 0.589 (m/s²), cao
M1 uses v² = u² + 2as with u = 0, s = 3.5, and their a
A1 v = awrt 2.03 (m/s), cao
Answer: a = awrt 0.589 m/s²; v = awrt 2.03 m/s

Question 19

M1 equation of motion for Q (down positive): 5g - T = 5a
M1 equation of motion for P: T - 0.25(2)(9.8) = 2a, i.e. T - 4.9 = 2a, using R = 2g for the normal reaction on the table
A1 adds the two equations to eliminate T and solves: a = 6.3 (m/s²), cao
A1 substitutes back to find T = 17.5 (N), cao (ft)
Answer: a = 6.3 m/s²; T = 17.5 N

Question 19

4 marks

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Mechanics: Projectiles and Applications Depth: Fluency and Exam Drill - Worksheets, Questions and Revision

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