A particle is projected horizontally with speed 12 m/s from the top of a cliff. State the initial vertical component of its velocity.
(Total for Question 1 is 1 mark)
2
A particle is projected with speed 30 m/s at an angle of 40 degrees above the horizontal. Find the initial vertical component of its velocity, giving your answer to 3 significant figures.
(Total for Question 2 is 1 mark)
3
A particle is projected with speed 30 m/s at an angle of 40 degrees above the horizontal. Find the initial horizontal component of its velocity, giving your answer to 3 significant figures.
(Total for Question 3 is 1 mark)
4
A ball is thrown horizontally from a height and takes 2 seconds to hit the ground. Using g = 9.8 m/s2, find the vertical component of its velocity at the instant it hits the ground.
(Total for Question 4 is 1 mark)
5
A particle is projected horizontally with speed 10 m/s from the top of a cliff 19.6 m above the sea. Using g = 9.8 m/s2, find the time taken for the particle to reach the sea.
(Total for Question 5 is 2 marks)
6
A particle is projected horizontally with speed 15 m/s from the top of a cliff. It takes 3 seconds to reach the sea. Find the horizontal distance travelled by the particle.
(Total for Question 6 is 2 marks)
7
A particle P has velocity (2i - j) m/s at time t = 0 and moves with constant acceleration (1i + 2j) m/s2. Find the velocity of P at time t = 4 seconds.
(Total for Question 7 is 2 marks)
8
A particle P has position vector (1i - 4j) m at time t = 0, velocity (3i + 0j) m/s at t = 0, and moves with constant acceleration (0i + 4j) m/s2. Find the position vector of P at time t = 2 seconds.
(Total for Question 8 is 2 marks)
9
A particle has position vector r = (t2 - 3t)i + (2t + 5)j metres at time t seconds. Find an expression for the velocity of the particle at time t.
(Total for Question 9 is 2 marks)
10
A particle moves with velocity v = 4ti - 3j m/s. Find the acceleration vector of the particle.
(Total for Question 10 is 1 mark)
11
A particle moves in a straight line with velocity v = 6t2 - 2 m/s at time t seconds. Find the acceleration of the particle when t = 3 seconds.
(Total for Question 11 is 2 marks)
12
A particle moves in a straight line with acceleration a = 4t - 1 m/s2 at time t seconds. Given that the particle is at rest when t = 0, find an expression for the velocity v m/s at time t.
(Total for Question 12 is 2 marks)
13
A particle is projected at an angle above horizontal ground and reaches its greatest height 2.5 seconds after projection. Using g = 9.8 m/s2 and the fact that the vertical velocity is zero at greatest height, find the initial vertical component of velocity.
(Total for Question 13 is 2 marks)
14
For a particle projected from and landing on horizontal ground, state the horizontal component of its velocity at the moment it lands, in terms of its initial horizontal component U.
(Total for Question 14 is 1 mark)
15
A particle is projected horizontally with speed 18 m/s from the top of a vertical cliff 78.4 m above the sea, modelled as a horizontal plane. Using g = 9.8 m/s2, find the horizontal distance from the base of the cliff to the point where the particle lands.
(Total for Question 15 is 3 marks)
16
A particle is projected from a point O on horizontal ground with speed 21 m/s at an angle of 40 degrees above the horizontal. Using g = 9.8 m/s2, find the greatest height reached by the particle above O, giving your answer to 3 significant figures.
(Total for Question 16 is 3 marks)
17
A particle is projected from a point O on horizontal ground with speed 24 m/s at an angle of 35 degrees above the horizontal. The particle moves freely under gravity until it returns to the ground. Using g = 9.8 m/s2, find the time of flight, and hence find the range of the particle. Give each answer to 3 significant figures.
(Total for Question 17 is 4 marks)
18
At time t = 0, a particle A has position vector (3i + 4j) m and moves with constant velocity (2i - j) m/s. At the same time, a particle B has position vector (-5i - 2j) m and moves with constant velocity (4i + 3j) m/s. Find the distance between A and B when t = 2 seconds, and state, with a reason, whether A and B collide at this time.
(Total for Question 18 is 4 marks)
19
A particle P of mass 0.4 kg moves under the action of a single force F newtons, where F = (3t + 2)i - 5j, and t is the time in seconds (t ≥ 0). When t = 0, P has velocity (1i + 2j) m/s. Find the velocity of P at time t = 3 seconds.
(Total for Question 19 is 4 marks)
Mark scheme · M4D Mechanics: Projectiles and Further Kinematics: Fluency and Exam Drill
Question 1
B1 0 m/s
Answer: 0 m/s
Question 2
B1 awrt 19.3 m/s
Answer: 19.3 m/s (3 s.f.)
Question 3
B1 awrt 23.0 m/s
Answer: 23.0 m/s (3 s.f.)
Question 4
B1 19.6 cao
Answer: 19.6 m/s
Question 5
M1 use s = 1/2 g t2 with s=19.6
A1 2 s cao
Answer: 2 s
Question 6
M1 horizontal distance = horizontal speed x time
A1 45 m cao
Answer: 45 m
Question 7
M1 v = u + at applied with t=4
A1 (6i + 7j) m/s cao
Answer: (6i + 7j) m/s
Question 8
M1 r = r0 + ut + 1/2 a t2 applied with t=2
A1 (7i + 4j) m cao
Answer: (7i + 4j) m
Question 9
M1 differentiate each component of r with respect to t
A1 v = (2t - 3)i + 2j cao
Answer: v = (2t - 3)i + 2j
Question 10
B1 4i cao
Answer: 4i m/s2 (i.e. 4i + 0j)
Question 11
M1 differentiate: a = 12t, substitute t=3
A1 36 m/s2 cao
Answer: 36 m/s2
Question 12
M1 integrate a with respect to t: v = 2t2 - t + C, use v(0)=0 to find C=0
A1 v = 2t2 - t cao
Answer: v = 2t2 - t
Question 13
M1 use 0 = u - gt with t=2.5, rearranged for u
A1 24.5 m/s cao
Answer: 24.5 m/s
Question 14
B1 it equals U, unchanged, since there is no horizontal acceleration
Answer: It equals U (unchanged).
Question 15
M1 find time of flight using s = 1/2 g t2 with s=78.4
M1 t = 4 s, then horizontal distance = 18 x 4
A1 72 m cao
Answer: 72 m
Question 16
M1 find initial vertical component: uy = 21 sin40
M1 use v2 = uy2 - 2 g h with v=0, rearranged for h
A1 awrt 9.30 m
Answer: 9.30 m (3 s.f.)
Question 17
M1 find uy = 24 sin35, use time of flight = 2 uy / g
A1 awrt 2.81 s
M1 find ux = 24 cos35, range = ux x time of flight (ft)
A1 awrt 55.2 m
Answer: time of flight = 2.81 s; range = 55.2 m (both 3 s.f.)
Question 18
M1 substitute t=2 into r = r0 + vt for both A and B
A1 A at (7i + 2j) m, B at (3i + 4j) m
M1 distance = magnitude of (position of A - position of B)
A1 awrt 4.47 m; they do not collide since the distance between them is not zero
Answer: Distance = 4.47 m (3 s.f.); A and B do not collide at t = 2
Question 19
M1 use F = ma to find a = F/m = (7.5t + 5)i - 12.5j
M1 integrate a with respect to t to find v(t), introducing a constant vector
A1 use v(0) = (1i + 2j) to find the constant, giving v = (3.75t2 + 5t + 1)i + (-12.5t + 2)j