A particle is projected horizontally from the top of a diving board. State the horizontal component of the particle's acceleration while it is in the air.
(Total for Question 1 is 1 mark)
2
A particle is projected at an angle above horizontal ground and moves freely under gravity. State the vertical component of the particle's acceleration during the flight, using g = 9.8 m/s2.
(Total for Question 2 is 1 mark)
3
A particle is projected from horizontal ground and moves freely under gravity, following a projectile path. State the direction of the particle's velocity at the instant it reaches its greatest height.
(Total for Question 3 is 1 mark)
4
A hang glider is launched from a hillside. The horizontal component of the hang glider's initial velocity is 12.4 m/s, at an angle of 37 degrees above the horizontal.
(a)Find the initial speed of the hang glider, giving your answer to 3 significant figures.(1)
(b)Find the initial vertical component of the hang glider's velocity, giving your answer to 3 significant figures.(1)
(Total for Question 4 is 2 marks)
5
A conker falls from rest from a window ledge 4.9 m above the ground. Using g = 9.8 m/s2 and modelling the conker as a particle, find the speed at which it hits the ground.
(Total for Question 5 is 2 marks)
6
A paraglider launches horizontally from a hillside and lands 37.4 m (measured horizontally) from the launch point after 3.4 seconds of flight. Find the horizontal component of the paraglider's velocity.
(Total for Question 6 is 2 marks)
7
A remote-controlled car P has velocity (4i - 9j) m/s at time t = 0 and moves with constant acceleration (-1i + 3j) m/s2. Find the value of t at which the velocity of P is parallel to the vector i.
(Total for Question 7 is 2 marks)
8
A particle has position vector r = (2t3 - 5t)i + (4t2 + 3)j metres at time t seconds. Find the velocity of the particle when t = 2 seconds.
(Total for Question 8 is 2 marks)
9
A particle P has velocity v = (4t - 1)i + 6j m/s at time t seconds. When t = 0, P has position vector (2i + 5j) m. Find the position vector of P at time t seconds.
(Total for Question 9 is 2 marks)
10
A particle P moves in a plane with acceleration a = (6t - 2)i + 4j m/s2 at time t seconds (t ≥ 0). When t = 0, P has velocity (1i - 3j) m/s and position vector (0i + 2j) m. Find the position vector of P at time t = 2 seconds.
(Total for Question 10 is 4 marks)
11
A particle is projected from horizontal ground with speed 24 m/s at an angle of 35 degrees above the horizontal. Using g = 9.8 m/s2, find the speed of the particle at time t = 1.5 seconds after projection, giving your answer to 3 significant figures.
(Total for Question 11 is 3 marks)
12
A particle is projected from horizontal ground with speed 18 m/s at an angle of 50 degrees above the horizontal. Using g = 9.8 m/s2, find the angle the particle's velocity makes with the horizontal at time t = 1 second after projection, giving your answer to 3 significant figures.
(Total for Question 12 is 3 marks)
13
A particle is projected from horizontal ground with speed 23 m/s at an angle of 44 degrees above the horizontal. Using g = 9.8 m/s2, show that the greatest height reached by the particle is 13.0 m, correct to 3 significant figures.
(Total for Question 13 is 3 marks)
14
A firework is launched from a platform 12 m above horizontal ground with initial speed 15 m/s at an angle of 25 degrees above the horizontal. Using g = 9.8 m/s2, find the time taken for the firework to reach the ground, giving your answer to 3 significant figures.
(Total for Question 14 is 3 marks)
15
A student was asked to find the horizontal component of the initial velocity of a particle projected at 30 m/s at an angle of 60 degrees above the horizontal. Their working is shown below:
Horizontal component = 30 sin60 = 26.0 m/s (3 s.f.)
Identify the error in the student's working, and find the correct value of the horizontal component, giving your answer to 3 significant figures.
(Total for Question 15 is 2 marks)
16
At time t = 0, a delivery drone A has position vector (2i + 6j) m and moves with constant velocity (3i - 1j) m/s. At the same time, a second drone B has position vector (10i + 2j) m and moves with constant velocity (-1i + 2j) m/s. Find the value of t at which the distance between A and B is least.
(Total for Question 16 is 3 marks)
17
A ball is thrown from a point on horizontal ground with speed 13 m/s and lands 15 m from the point of projection. Using the range formula R = u2 sin(2*θ) / g and taking g = 9.8 m/s2, find the smaller possible angle of projection θ, giving your answer to 3 significant figures.
(Total for Question 17 is 2 marks)
18
A ball is thrown from a point 1.5 m above horizontal ground with speed 20 m/s at an angle of 15 degrees above the horizontal. Using g = 9.8 m/s2:
(a)Find the time taken for the ball to reach the ground, giving your answer to 3 significant figures.(2)
(b)Find the horizontal distance travelled by the ball before it lands, giving your answer to 3 significant figures. Use your unrounded value from part (a).(2)
(Total for Question 18 is 4 marks)
19
A particle P of mass 0.5 kg is acted on by a constant force F = (6i - 4j) N. When t = 0, P is at rest at the origin. Find the position vector of P at time t = 4 seconds.
(Total for Question 19 is 3 marks)
Mark scheme · M4DB Mechanics: Projectiles and Further Kinematics: Fluency and Exam Drill (Part 2)
Answer: 19.0 degrees above the horizontal (3 s.f.)
Question 13
M1 find uy = 23sin44 (= awrt 16.0)
M1 use v2 = uy2 - 2gh with v=0, rearranged for h
A1 h = 13.0 m (3 s.f.) correctly shown, cso
Answer: Greatest height = 13.0 m (as required)
Question 14
M1 find uy = 15sin25 and form -12 = uy t - 4.9t2, rearranged to 4.9t2 - 6.34t - 12 = 0 oe
M1 solve the quadratic using the quadratic formula, selecting the positive root
A1 awrt 2.34 s
Answer: 2.34 s (3 s.f.)
Question 15
M1 identify the error: the student used sin60 (which gives the vertical component) instead of cos60 for the horizontal component
A1 correct value: 15.0 m/s cao
Answer: Error: sin60 was used instead of cos60 (sin60 gives the vertical component, not the horizontal one). Correct horizontal component = 30cos60 = 15.0 m/s.
Question 16
M1 find the position vector of A relative to B as a function of t: (4t-8)i + (4-3t)j oe
M1 form (distance)2 = (4t-8)2 + (4-3t)2 = 25t2 - 88t + 80 and minimise (differentiate and set to 0, or use t=-b/2a)
A1 t = 1.76 s cao
Answer: t = 1.76 s
Question 17
M1 substitute into R=u2 sin(2*θ)/g: 15 = 169 sin(2*θ)/9.8, rearrange to sin(2*θ) = 147/169 (awrt 0.870), and solve for 2*θ using arcsin (taking the value less than 90 degrees)
A1 θ = awrt 30.2 degrees
Answer: θ = 30.2 degrees (3 s.f.)
Question 18
(a) M1 find uy=20sin15 and set up and solve 4.9t2 - 5.18t - 1.5 = 0 using the quadratic formula, selecting the positive root
(a) A1 awrt 1.29 s
(a) Answer: 1.29 s (3 s.f.)
(b) M1 horizontal distance = 20cos15 x t (ft their time from part (a))