Kinematics of a particle in a straight line or plane — WME02 IAL Mathematics practice questions · Study-IAL

Kinematics of a particle in a straight line or plane

WME02Topic 2.13 questions

Projectiles and motion with variable acceleration using calculus and vectors.

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Question 16 marks
A particle $P$ moves along the $x$ -axis. At time $t$ seconds ( $t \ge 0$ ) its velocity is $v = (2t^2 - 9t + 4)\,\text{m s}^{-1}$ . When $t = 0$ , $P$ is at the origin.

(a) Find the values of $t$ for which $P$ is instantaneously at rest.

(b) Find the total distance travelled by $P$ between $t = 0$ and $t = 4$ .
Mark scheme
- (a) M1 Solve $2t^2 - 9t + 4 = 0$ , i.e. $(2t - 1)(t - 4) = 0$ (1)
Question 27 marks
A ball is projected from a point $O$ on horizontal ground with speed $21\,\text{m s}^{-1}$ at an angle $\alpha$ above the horizontal, where $\tan\alpha = \tfrac{4}{3}$ . The ball is modelled as a particle moving freely under gravity.

−

1)(t−

4)=

t = 0.5

and

t = 4

(1)

(b) M1

x = \tfrac{2}{3}t^3 - \tfrac{9}{2}t^2 + 4t

(no constant as

x = 0

t = 0

) (1)

x(0.5) = \tfrac{23}{24} = 0.958\,\text{m}

(1)

x(4) = -\tfrac{40}{3} = -13.3\,\text{m}

(1)

A1 Distance

= \tfrac{23}{24} + \left(\tfrac{23}{24} + \tfrac{40}{3}\right) = 15.25\,\text{m}

(1)

tan α = \frac{4}{3}

(a) Find the greatest height of the ball above the ground.

(b) Find the length of time for which the ball is at least $10\,\text{m}$ above the ground.

Mark scheme

(a) B1 Initial components: horizontal $21 \times 0.6 = 12.6$ , vertical $21 \times 0.8 = 16.8\,\text{m s}^{-1}$ (1)
A1 $0 = 16.8^2 - 2 \times 9.8 \times h$ , so $h = 14.4\,\text{m}$ (1)
(b) M1 $10 = 16.8t - 4.9t^2$ (1)
A1 $t = 0.767$ or $t = 2.66$ (1)
A1 Time $= 2.66 - 0.767 = 1.9\,\text{s}$ (1)
(c) M1 Time of flight: $0 = 16.8t - 4.9t^2$ , so $t = 3.43\,\text{s}$ (1)
A1 Range $= 12.6 \times 3.43 = 43.2\,\text{m}$ (1)