Edexcel A level Maths (9MA0) · Mechanics › Kinematics
Practise Projectiles. 1 exam-style questions plus unlimited generated ones on this subtopic, at up to four difficulty levels, with full mark schemes and a progress tracker. Free, no account needed.
Written for this site in the style of Edexcel exam questions. They are not taken from real past papers.
Question 1Easy6 marks
[Take \(g = 9.8\) m s−2. Give any answer that uses this value of \(g\) to 2 or 3 significant figures.] A small stone is thrown horizontally with speed 6 m s−1 from a point \(A\) at the top of a vertical cliff; \(A\) is 44.1 m above the sea. The small stone hits the sea at the point \(B\). The small stone is modelled as a particle moving freely under gravity.
[object Object]
(a) Find the time taken for the small stone to travel from \(A\) to \(B\).[3]
(b) Find the horizontal distance travelled by the small stone from \(A\) to \(B\).[2]
(c) State one limitation of the model.[1]
Show the answer and mark scheme
(a)Answer: \(3\) s
M1 for using vertical motion with zero initial vertical velocity
A1 for \(44.1 = \tfrac{1}{2} \times 9.8 \times t^2\)
A1 for 3 (s)
Worked solution: Vertically: \(u = 0, \ a = 9.8, \ s = 44.1\), so \(44.1 = 4.9t^2 \Rightarrow t = \sqrt{\frac{44.1}{4.9}} = 3\) s
(b)Answer: \(18\) m
M1 for \(6 \times \text{their } t\)
A1ft for 18 (m)
Worked solution: Horizontally the speed is constant: \(6 \times 3 = 18\) m
(c)Answer: Air resistance is ignored; or the small stone is not a particle (its size and any spin are ignored).
B1 for air resistance ignored / the small stone has size or spins / \(g\) is taken as constant
Worked solution: For example, air resistance on the small stone has been ignored.
Question 2Medium9 marks
[Take \(g = 9.8\) m s−2. Give any answer that uses this value of \(g\) to 2 or 3 significant figures.] A marble rolls off the edge of a horizontal table top with speed 2 m s−1. The table top is 0.784 m above a horizontal floor and the marble leaves the table at the point \(A\). The marble hits the floor at the point \(B\). The marble is modelled as a particle moving freely under gravity.
[object Object]
(a) Find the time taken for the marble to travel from \(A\) to \(B\).[2]
(b) Find the horizontal distance travelled by the marble from \(A\) to \(B\).[2]
(c) Find the speed of the marble as it reaches \(B\).[3]
(d) Find the angle between the direction of motion of the marble and the horizontal as it reaches \(B\). Give your answer in degrees to one decimal place.[2]
Show the answer and mark scheme
(a)Answer: \(0.4\) s
M1 for \(0.784 = \tfrac{1}{2} \times 9.8 \times t^2\)
A1 for 0.4 (s)
Worked solution: Vertically: \(u = 0, \ a = 9.8, \ s = 0.784\), so \(0.784 = 4.9t^2 \Rightarrow t = \sqrt{\frac{0.784}{4.9}} = 0.4\) s
(b)Answer: \(0.8\) m
M1 for \(2 \times \text{their } t\)
A1ft for 0.8 (m)
Worked solution: Horizontally the speed is constant: \(2 \times 0.4 = 0.8\) m
(c)Answer: \(4.40\) m s−1 (accept \(4.4\) m s−1) m s−1
M1 for the vertical component of the velocity at \(B\): \(9.8 \times \text{their } t\) or \(\sqrt{2 \times 9.8 \times 0.784}\)
M1 for combining with the horizontal component 2 using Pythagoras
A1 for 4.40 or 4.4 (m s−1)
Worked solution: At \(B\): horizontal component 2 m s−1; vertical component \(9.8 \times 0.4 = 3.92\) m s−1 (downwards). \(\text{speed} = \sqrt{2^2 + 3.92^2} = 4.4007\ldots = 4.40\) m s−1
(d)Answer: \(63.0^\circ\) below the horizontal °
M1 for \(\tan\theta = \frac{\text{vertical component}}{\text{horizontal component}}\) with their values
A1 for \(63.0^\circ\)
Worked solution: \(\tan\theta = \frac{3.92}{2} \Rightarrow \theta = 63.0^\circ\) below the horizontal.
Question 3Hard11 marks
[Take \(g = 9.8\) m s−2. Give any answer that uses this value of \(g\) to 2 or 3 significant figures.] A ball is thrown horizontally from a point \(A\) which is 78.4 m above horizontal ground. The ball moves freely under gravity and hits the ground at the point \(B\), where the horizontal distance of \(B\) from \(A\) is 56 m.
[object Object]
(a) Find the time taken for the ball to travel from \(A\) to \(B\).[2]
(b) Find the speed with which the ball was projected.[2]
(c) Find the speed of the ball when it hits the ground.[3]
(d) Find the height of the ball above the ground at the instant when its direction of motion makes an angle of \(45^\circ\) with the horizontal.[4]
Show the answer and mark scheme
(a)Answer: \(4\) s
M1 for \(78.4 = \tfrac{1}{2} \times 9.8 \times t^2\)
A1 for 4 (s)
Worked solution: \(78.4 = 4.9t^2 \Rightarrow t^2 = 16 \Rightarrow t = 4\) s
(b)Answer: \(14\) m s−1
M1 for \(u \times \text{their } t = 56\)
A1ft for 14 (m s−1)
Worked solution: \(u \times 4 = 56 \Rightarrow u = 14\) m s−1
(c)Answer: \(41.6\) m s−1 (accept \(42\) m s−1) m s−1
M1 for the vertical component \(9.8 \times 4 = 39.2\)
M1 for Pythagoras with the horizontal component
A1 for 41.6 or 42 (m s−1)
Worked solution: \(\sqrt{14^2 + 39.2^2} = 41.6249\ldots = 41.6\) m s−1
(d)Answer: \(68.4\) m (accept \(68\) m) m
M1 for the vertical component of velocity equal to the horizontal component: \(9.8t = 14\)
A1 for \(t = 1.43\)
M1 for the distance fallen \(4.9t^2\) subtracted from 78.4
A1 for 68.4 or 68 (m)
Worked solution: At \(45^\circ\) the vertical and horizontal components of velocity are equal: \(9.8t = 14 \Rightarrow t = 1.4285\ldots = 1.43\) s. Height \(= 78.4 - 4.9 \times 1.4285\ldots^2 = 68.4\) m