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A-Level · Mechanics

M4 Mechanics: Projectiles and Further Kinematics

EDEXCEL 9MA0 · Calculator allowed · about 180 minutes

Total Marks

Name: _______________________________ Date: ____ / ____ / ______

Answer ALL questions. Show all your working.

A small stone is thrown horizontally with speed 15 m/s from the top of a vertical cliff. The point of projection is 44.1 m vertically above the sea, which may be modelled as a horizontal plane. The stone is modelled as a particle moving freely under gravity. Use g = 9.8 m/s^2.

(a)Show that the time taken for the stone to reach the sea is exactly 3 seconds.(3)

(b)Find the horizontal distance from the base of the cliff to the point where the stone lands.(2)

(c)Find the speed of the stone as it hits the sea, giving your answer to 3 significant figures.(3)

(d)Find the angle the velocity makes with the horizontal as the stone hits the sea, giving your answer to the nearest degree.(2)

(Total for Question 1 is 10 marks)

A particle is projected horizontally with speed U m/s from a point on top of a vertical cliff. The point of projection is 78.4 m above the sea, modelled as a horizontal plane, and the particle lands in the sea at a horizontal distance of 98 m from the base of the cliff. The particle is modelled as moving freely under gravity. Use g = 9.8 m/s^2.

(a)Find the time taken for the particle to reach the sea.(3)

(b)Find the value of U.(2)

(c)Find the speed of the particle immediately before it enters the sea, giving your answer to 3 significant figures.(3)

(d)Find the angle the velocity makes with the horizontal immediately before impact, to the nearest degree.(2)

(Total for Question 2 is 10 marks)

At time t = 0, a particle P has position vector (2i - 3j) m relative to a fixed origin O, and velocity (4i + j) m/s. P moves with constant acceleration (-i + 2j) m/s^2. Position vectors are measured in metres and time in seconds.

(a)Find the velocity of P at time t = 3 seconds.(2)

(b)Find the position vector of P at time t = 3 seconds.(3)

(c)Find the value of t for which P is moving parallel to the vector (i + j).(3)

(d)Find the speed of P at the instant found in part (c).(2)

(Total for Question 3 is 10 marks)

A particle is projected from a point O on horizontal ground with speed 28 m/s at an angle of 30 degrees above the horizontal. The particle moves freely under gravity until it returns to the horizontal ground. Use g = 9.8 m/s^2.

(a)Find the time taken for the particle to return to the ground.(3)

(b)Find the greatest height reached by the particle above the ground.(3)

(c)Find the horizontal distance travelled by the particle before it first returns to the ground (the range).(3)

(d)Show that, for a particle projected with speed u at angle θ above horizontal ground and landing at the same level, the range is given by R = u² sin(2theta)/g.(4)

(Total for Question 4 is 13 marks)

A particle P moves in a plane so that at time t seconds (t ≥ 0), its velocity is v = (3t² - 6)i + (4t - 2)j m/s. When t = 0, P is at the point with position vector (5i + 2j) m relative to a fixed origin O.

(a)Find an expression for the acceleration of P at time t seconds, and hence find the acceleration when t = 2.(3)

(b)Find the position vector of P at time t = 2 seconds.(4)

(c)Show that P is moving parallel to the vector i when t = 0.5 seconds.(2)

(d)Find the speed of P at this instant.(2)

(Total for Question 5 is 11 marks)

A particle P moves along the positive x-axis. At time t seconds (0 ≤ t ≤ 4), the velocity of P is v m/s, where v = 12t - 3t^2. When t = 0, P is at the origin O.

(a)Find the acceleration of P when t = 1.(2)

(b)Find the value of t, other than t = 0, at which P is instantaneously at rest.(2)

(c)Find the distance travelled by P between t = 0 and t = 4.(4)

(d)Find the maximum speed of P for 0 ≤ t ≤ 4.(3)

(Total for Question 6 is 11 marks)

A particle is projected from a point A at the top of a vertical cliff with speed 32 m/s at an angle of 25 degrees above the horizontal, in a direction away from the cliff. The point A is 55 m above the sea, modelled as a horizontal plane. The particle moves freely under gravity until it enters the sea. Use g = 9.8 m/s^2.

(a)Find the initial horizontal and vertical components of the velocity, each to 3 significant figures.(2)

(b)Find the time taken for the particle to enter the sea, giving your answer to 3 significant figures.(4)

(c)Find the horizontal distance from the base of the cliff to the point where the particle enters the sea, to 3 significant figures.(2)

(d)Find the speed of the particle as it enters the sea, to 3 significant figures.(3)

(Total for Question 7 is 11 marks)

A particle P of mass 0.5 kg moves under the action of a single force F newtons, where F = (6t - 3)i + 4j, and t is the time in seconds (t ≥ 0). When t = 0, P has velocity (2i - j) m/s.

(a)Using F = ma, find an expression for the acceleration of P at time t seconds.(2)

(b)Show that the velocity of P at time t seconds is given by v = (6t² - 6t + 2)i + (8t - 1)j.(3)

(c)Find the value of t at which P is moving parallel to the vector i.(3)

(d)Find the acceleration of P at this instant, and hence find the magnitude of the force acting on P at this time, giving your answer to 3 significant figures.(3)

(Total for Question 8 is 11 marks)

At time t = 0, a particle A has position vector (4i - 2j) m relative to a fixed origin O, and moves with constant velocity (2i + 3j) m/s. At the same time, a particle B has position vector (-6i + 10j) m and moves with constant velocity (5i - j) m/s. Positions are measured in metres and time in seconds.

(a)Find the position vector of A at time t seconds.(2)

(b)Find the position vector of B at time t seconds.(2)

(c)Show that d² = 25t² - 156t + 244, where d metres is the distance between A and B at time t seconds.(3)

(d)Hence find the minimum distance between A and B, and the value of t at which this occurs.(4)

(Total for Question 9 is 11 marks)

A particle is projected from a point O on horizontal ground with speed U m/s at an angle α above the horizontal, where tan(α) = 3/4. Given that U = 35 m/s, use g = 9.8 m/s² throughout.

(a)Show that the initial horizontal and vertical components of the velocity are 28 m/s and 21 m/s respectively.(2)

(b)Find the time taken for the particle to reach its greatest height.(2)

(c)Find the greatest height above O reached by the particle.(3)

(d)Find the range of the projectile on the horizontal ground through O.(3)

(e)A vertical wall of height 15 m stands on the ground, in the vertical plane of the particle's motion, at a horizontal distance of 80 m from O. Determine, showing your working, whether the particle passes over the wall, strikes the wall, or lands before reaching the wall.(5)

(Total for Question 10 is 15 marks)

A particle is projected from a point O on horizontal ground with speed 20 m/s at an angle θ above the horizontal, where 0 < θ < 90. Relative to O, the particle passes through the point with horizontal distance 15 m and height 5 m above the ground. Use g = 9.8 m/s^2.

(a)State the equation of the trajectory of a projectile launched from the origin with speed u at angle θ to the horizontal, in terms of x, y, u, θ and g.(2)

(b)Given that u = 20 m/s and the particle passes through the point (15, 5), show that 441 tan²(θ) - 2400 tan(θ) + 1241 = 0.(4)

(c)Hence find the two possible values of θ, giving each to 1 decimal place.(5)

(Total for Question 11 is 11 marks)

A particle is projected from a point O with speed 40 m/s at an angle of 50 degrees above the horizontal. The point O lies on a straight slope which is inclined at 20 degrees to the horizontal and rises in the same vertical plane as the motion of the particle. The particle lands at the point A on the slope. Horizontal and vertical distances from O are denoted x and y respectively (both measured in metres). Use g = 9.8 m/s^2.

(a)Write down the equation of the slope in terms of x and y, and the equation of the trajectory of the particle in terms of x and y.(3)

(b)Hence find the horizontal distance travelled by the particle before it lands at A, giving your answer to 3 significant figures.(5)

(c)Find the distance OA.(3)

(d)Find the time of flight from O to A, giving your answer to 3 significant figures.(3)

(Total for Question 12 is 14 marks)

Mark scheme · M4 Mechanics: Projectiles and Further Kinematics

Question 1

(a) M1 use s = ut + 0.5at² vertically with u = 0, a = 9.8, s = 44.1 (oe)
(a) M1 rearrange to t² = 9
(a) A1 t = 3 (cso)
(a) Answer: t = 3 s
(b) M1 horizontal distance = 15 x 3 (oe)
(b) A1 45 m (cao)
(b) Answer: 45 m
(c) M1 vertical speed v_y = gt = 9.8 x 3 = 29.4 (oe)
(c) M1 combine components: speed = √15² + 29.4²
(c) A1 awrt 33.0 m/s
(c) Answer: 33.0 m/s (3sf)
(d) M1 tan(θ) = 29.4/15 (oe)
(d) A1 θ = awrt 63 degrees
(d) Answer: 63 degrees below the horizontal (nearest degree)

Question 2

(a) M1 use s = ut + 0.5at² vertically with u = 0, s = 78.4 (oe)
(a) M1 rearrange to t² = 16
(a) A1 t = 4 (cao)
(a) Answer: 4 s
(b) M1 U = 98/4 (oe)
(b) A1 U = 24.5 m/s (cao)
(b) Answer: 24.5 m/s
(c) M1 v_y = gt = 9.8 x 4 = 39.2 (oe)
(c) M1 speed = √24.5² + 39.2²
(c) A1 awrt 46.2 m/s
(c) Answer: 46.2 m/s (3sf)
(d) M1 tan(θ) = 39.2/24.5 = 1.6 (oe)
(d) A1 θ = awrt 58 degrees
(d) Answer: 58 degrees below the horizontal

Question 3

(a) M1 use v = u + at with correct substitution
(a) A1 v = (i + 7j) m/s (cao)
(a) Answer: (i + 7j) m/s
(b) M1 use r = r0 + ut + 0.5at²
(b) M1 correct substitution of all vectors and t = 3
(b) A1 r = (9.5i + 9j) m (cao)
(b) Answer: (9.5i + 9j) m
(c) M1 v(t) = (4-t)i + (1+2t)j
(c) M1 equate components: 4 - t = 1 + 2t (oe)
(c) A1 t = 1 (cao)
(c) Answer: t = 1 s
(d) M1 v = 3i + 3j, speed = √3² + 3²
(d) A1 3sqrt(2) m/s, awrt 4.24 m/s
(d) Answer: 3sqrt(2) m/s = awrt 4.24 m/s

Question 4

(a) M1 initial vertical speed = 28 sin(30) = 14 (oe)
(a) M1 T = 2 x 14 / 9.8 (oe)
(a) A1 T = 20/7 s = awrt 2.86 s
(a) Answer: T = 20/7 s = awrt 2.86 s
(b) M1 use v² = u² - 2gs with v = 0, u = 14
(b) M1 rearrange for s
(b) A1 H = 10 m (cao)
(b) Answer: 10 m
(c) M1 horizontal speed = 28 cos(30) (oe)
(c) M1 R = horizontal speed x T
(c) A1 R = 40sqrt(3) m, awrt 69.3 m
(c) Answer: 40sqrt(3) m = awrt 69.3 m
(d) M1 time of flight T = 2u sin(θ)/g stated or derived
(d) M1 R = (u cos(θ)) x T
(d) M1 substitute and use sin(2theta) = 2 sin(θ) cos(θ)
(d) A1 R = u² sin(2theta)/g (cso, no errors seen)
(d) Answer: R = u² sin(2theta)/g (shown)

Question 5

(a) M1 differentiate v to obtain a = 6ti + 4j
(a) A1 correct expression for a(t)
(a) A1 a(2) = 12i + 4j m/s² (cao)
(a) Answer: a = 6ti + 4j; at t = 2, a = 12i + 4j m/s²
(b) M1 integrate v(t) componentwise, including constants of integration
(b) A1 i-component t³ - 6t + C1
(b) A1 j-component 2t² - 2t + C2
(b) A1 use r(0) = 5i + 2j to find constants, giving r(2) = (i + 6j) m (cao)
(b) Answer: (i + 6j) m
(c) M1 set j-component of v to zero: 4t - 2 = 0
(c) A1 t = 0.5 (cso)
(c) Answer: t = 0.5 s
(d) M1 v(0.5) = (3(0.5)² - 6)i + 0j = -5.25i
(d) A1 speed = 5.25 m/s (cao)
(d) Answer: 5.25 m/s

Question 6

(a) M1 a = dv/dt = 12 - 6t
(a) A1 a(1) = 6 m/s² (cao)
(a) Answer: 6 m/s²
(b) M1 set v = 0: 12t - 3t² = 0, factorise (oe)
(b) A1 t = 4 (cao)
(b) Answer: t = 4 s
(c) M1 recognise v ≥ 0 throughout so distance = displacement
(c) M1 integrate: integral of (12t - 3t²) dt = 6t² - t³
(c) M1 substitute limits t = 0 and t = 4
(c) A1 distance = 32 m (cao)
(c) Answer: 32 m
(d) M1 set a = 0: 12 - 6t = 0, t = 2
(d) M1 justify maximum (a changes from positive to negative, oe)
(d) A1 v(2) = 12 m/s (cao)
(d) Answer: 12 m/s

Question 7

(a) M1 u_x = 32cos(25), u_y = 32sin(25) identified
(a) A1 u_x = awrt 29.0 m/s and u_y = awrt 13.5 m/s
(a) Answer: u_x = 29.0 m/s, u_y = 13.5 m/s
(b) M1 form vertical equation -55 = 13.524t - 4.9t² (oe, using unrounded u_y)
(b) M1 rearrange to 4.9t² - 13.524t - 55 = 0 (oe)
(b) M1 solve quadratic (formula), reject negative root
(b) A1 t = awrt 5.00 s
(b) Answer: t = awrt 5.00 s
(c) M1 x = u_x x t (ft from (a) and (b))
(c) A1 x = awrt 145 m
(c) Answer: awrt 145 m
(d) M1 vertical speed at impact v_y = √u_y² + 2 x 9.8 x 55 (oe, or v_y = u_y - gt)
(d) M1 combine: speed = √u_x² + v_y²
(d) A1 speed = awrt 45.8 m/s
(d) Answer: awrt 45.8 m/s

Question 8

(a) M1 a = F/m
(a) A1 a = (12t - 6)i + 8j m/s² (cao)
(a) Answer: a = (12t - 6)i + 8j m/s²
(b) M1 integrate a(t) to find v(t) with a constant vector
(b) A1 correct unsimplified expression, e.g. (6t²-6t+C1)i + (8t+C2)j
(b) A1 use v(0) = 2i - j to find constants and complete (cso)
(b) Answer: v = (6t² - 6t + 2)i + (8t - 1)j (shown)
(c) M1 set j-component of v to zero: 8t - 1 = 0
(c) M1 solve for t
(c) A1 t = 0.125 s (= 1/8 s, cao)
(c) Answer: t = 0.125 s
(d) M1 a(0.125) = (12x0.125-6)i + 8j = -4.5i + 8j
(d) M1 F = ma, magnitude = 0.5 x √4.5² + 8²
(d) A1 F = awrt 4.59 N
(d) Answer: awrt 4.59 N

Question 9

(a) M1 r_A = r0 + vt substituted
(a) A1 r_A = (4+2t)i + (-2+3t)j
(a) Answer: (4+2t)i + (-2+3t)j m
(b) M1 r_B = r0 + vt substituted
(b) A1 r_B = (-6+5t)i + (10-t)j
(b) Answer: (-6+5t)i + (10-t)j m
(c) M1 find vector AB = r_B - r_A
(c) M1 form d² = (i-component)² + (j-component)²
(c) A1 d² = 25t² - 156t + 244 (cso)
(c) Answer: d² = 25t² - 156t + 244 (shown)
(d) M1 differentiate d² with respect to t and set equal to zero: 50t - 156 = 0
(d) A1 t = 3.12 s (cao)
(d) M1 substitute back to find d²_min
(d) A1 d_min = 0.8 m (cao)
(d) Answer: t = 3.12 s, minimum distance = 0.8 m

Question 10

(a) M1 use sin(α) = 3/5, cos(α) = 4/5 (from tan(α) = 3/4) with U = 35
(a) A1 both components shown: 28 m/s and 21 m/s (cso)
(a) Answer: horizontal = 28 m/s, vertical = 21 m/s (shown)
(b) M1 t = 21/9.8
(b) A1 t = 15/7 s = awrt 2.14 s
(b) Answer: 15/7 s = awrt 2.14 s
(c) M1 use v² = u² - 2gs with v = 0, u = 21
(c) M1 rearrange for s
(c) A1 H = 22.5 m (cao)
(c) Answer: 22.5 m
(d) M1 time of flight T = 2 x 15/7 = 30/7
(d) M1 R = 28 x T
(d) A1 R = 120 m (cao)
(d) Answer: 120 m
(e) M1 find time to reach horizontal distance 80: t = 80/28 = 20/7
(e) M1 substitute into vertical displacement equation y = 21t - 4.9t²
(e) A1 y = 20 m at x = 80 m
(e) M1 compare 20 m with wall height 15 m and confirm 80 m < range 120 m
(e) A1 correct conclusion: the particle clears the wall, passing 5 m above it (cao)
(e) Answer: The particle passes over the wall (height at x=80 m is 20 m, greater than the 15 m wall; 80 m is less than the range of 120 m)

Question 11

(a) B1 y = x tan(θ) - ... term with correct x tan(θ) component
(a) B1 correct full trajectory equation y = x tan(θ) - (g x² (1+tan²(θ)))/(2u²) (oe, e.g. using sec²(θ))
(a) Answer: y = x tan(θ) - (g x² (1 + tan²(θ))) / (2u²)
(b) M1 substitute x=15, y=5, u=20, g=9.8 into trajectory equation
(b) M1 use 1+tan²(θ) = sec²(θ) and simplify the coefficient 9.8x225/(2x400) = 2.75625 (oe)
(b) M1 rearrange into a quadratic in tan(θ) and clear decimals
(b) A1 441 tan²(θ) - 2400 tan(θ) + 1241 = 0 (cso, or exact equivalent unsimplified form with correct working)
(b) Answer: 441 tan²(θ) - 2400 tan(θ) + 1241 = 0 (shown)
(c) M1 use the quadratic formula to solve for tan(θ)
(c) A1 tan(θ) = awrt 4.86 or 0.579 (both values, ft from (b))
(c) M1 use arctan to find θ for each value
(c) A1 θ = awrt 78.4 degrees
(c) A1 θ = awrt 30.1 degrees
(c) Answer: θ = awrt 30.1 degrees or θ = awrt 78.4 degrees

Question 12

(a) B1 slope equation y = x tan(20) (oe)
(a) B1 trajectory equation with θ=50, u=40 substituted: y = x tan(50) - (9.8x²(1+tan²(50)))/(2x40²) (oe)
(a) B1 correctly simplified coefficient (accept unsimplified equivalent form)
(a) Answer: y = x tan(20); y = x tan(50) - (9.8 x² sec²(50)) / 3200
(b) M1 equate the two equations for y
(b) M1 factor out x (x not equal to 0) to obtain a linear equation in x
(b) M1 substitute numerical values of tan(20), tan(50) and sec²(50)
(b) A1 correct simplified linear equation, e.g. 0.0074121x = 0.827784 (oe)
(b) A1 x = awrt 112 m
(b) Answer: x = awrt 112 m
(c) M1 OA = x / cos(20) (oe, or use Pythagoras with y = x tan(20))
(c) A1 correct method with ft value of x from (b)
(c) A1 OA = awrt 119 m
(c) Answer: awrt 119 m
(d) M1 use x = u cos(50) x t
(d) M1 rearrange for t, ft value of x from (b)
(d) A1 t = awrt 4.34 s
(d) Answer: awrt 4.34 s

Mechanics: Projectiles and Further Kinematics - Worksheets, Questions and Revision