Edexcel A-Level Mathematics AS M1.3 Kinematics of a Particle Moving in a Straight Line Questions

Straight-line kinematics questions usually turn on choosing the right model: suvat for constant acceleration, graph gradients for acceleration and areas for distance.

Exam points

Choose a suvat equation for time, speed, height or displacement in vertical motion.
Use speed-time graph gradients for acceleration and areas for distance travelled.
Equate displacement expressions when two particles meet or one vehicle catches another.

Covered syllabus points

M1.3.1 - Constant-acceleration motion in a straight line

[Maximum number: 5]

Figure 3

A particle A of mass 4 kg is held at rest on a rough horizontal table. Particle A is attached to one end of a string that passes over a pulley P. The pulley is fixed the the of the table. The other end of the string is attached to a particle B, of mass 3 kg, which hangs freely below P.

The part of the string from A to P is perpendicular to the edge of the table and A, P and B all lie in the same vertical plane.

The string is modelled as being light and inextensible and the pulley is modelled as being small, smooth and light.

The system is released from rest with the string taut. At the instant of release, A is 2 m from the edge of the table and B is 1.4 m above a horizontal floor, as shown in Figure 3.

After descending with constant acceleration for 2 seconds, B hits the floor and does not rebound.

Determine, by calculation, whether or $\operatorname{not} A$ reaches the pulley.

$v^2=2\times0.7\times1.4$ $-\mu\times4g=4a$ $0^2=1.4^2-2\times\frac{5g}{8}s$ $s=0.16\text{ or }0.159$ 0.16+1.4<2

Therefore A does not reach the pulley.

M1 M1 M1 A1 A1 cso

Alternative for the final 3 marks:
$v^2=1.4^2-2\times\frac{5g}{8}\times0.6$ $v^2=-5.39\text{ or }-5.4488$

Since $v^2\ge0$ , A does not reach the pulley.

Marking guidance for this part:
M1 for finding the speed or speed squared of either particle when B hits the floor.
M1 for an equation of motion for A, allowing without the negative sign.
M1 for a complete method to find the distance moved by A until it stops;
condone sign error. This independent M mark is M0 if no new deceleration has been found.
A1 for the correct distance.
A1 cso for the correct conclusion, correctly reached, with <2 seen or 2 used in the working.
Alternative: M1 for a complete method to find $v^2$ for the speed with which it would hit the pulley;
A1 for the correct value of $v^2$ ;
A1 cso for the conclusion.

[Maximum number: 3]

Figure 5

A parcel of mass 2 kg is pulled up a rough inclined plane by the action of a constant force.

The force has magnitude 18 N and acts at an angle of 40^ to the plane.
The line of action of the force lies in a vertical plane containing a line of greatest slope of the inclined plane.

The plane is inclined at an angle of 30^ to the horizontal, as shown in Figure 5.
The coefficient of friction between the plane and the parcel is 0.3
The parcel is modelled as a particle P

The points A and B lie on a line of greatest slope of the plane, where A B=5 m and B is above A. Particle P passes through A with speed 2 ms^-1 in the direction AB.

The force of 18 N is removed at the instant P passes through B. As a result, P comes to rest at the point C.

Find the speed of P as it passes through B.

$v^2=2^2+2(1.18)(5)$
M1 A1ft

$v=3.98\text{ or }4.0\text{ or }4\text{ m s}^{-1}$
A1 cao

N.B. For (a) and (b), penalise over-accurate answers once only.

[Maximum number: 12]

A particle is projected vertically upwards from a point A with speed $24 \mathrm{~m} \mathrm{~s}^{-1}$

The point A is 2.5 m vertically above the point B.
Point B lies on horizontal ground.
The particle moves freely under gravity until it hits the ground at B with speed $V \mathrm{~m} \mathrm{~s}^{-1}$ After hitting the ground the particle does not rebound.

(a)

A to $B: \quad V^{2}=24^{2}+2(-g)(-2.5)$

M1 A1

OR: e.g. $0=24^{2}-2 g h$ and $V^{2}=2 g(h+2.5)$ oe

V=25

A1

(3)

(a)

Find the value of V.

[ 3 ]

A to $B: \quad V^{2}=24^{2}+2(-g)(-2.5)$
OR: e.g. $0=24^{2}-2 g h$ and $V^{2}=2 g(h+2.5)$ oe

V=25
(3)

(b)

Find the time taken for the particle to reach B.

The point C is 10 m vertically above A.

[ 3 ]

Some possible equations in t:
$\begin{aligned} 25&=-24+gt\\ 2.5&=\frac{(25+(-24))t}{2}\\ 2.5&=-24t+\frac12gt^2\\ 2.5&=25t-\frac12gt^2 \end{aligned}$
Or find $t_{\mathrm{UP}}=\frac{24}{g}$ and $t_{\mathrm{DOWN}}=\frac{25}{g}$ , then add.
$t=5\text{ s}$
(3)

(c)

Find the length of time for which the particle is above C.

[ 4 ]

From A to $C: 10=24 t+\frac{1}{2}(-g) t^{2}$
Complete method to find the required time: e.g. solving the above quadratic and finding the positive difference in the roots

N.B. Allow this mark if they solve their quadratic, and give the answer as a range of values: $t_{1},, t_{,,} t_{2}$
4, 4.0 or 3.98 (s)
ALT 1: From A to $C: W^{2}=24^{2}-2 \times 10 g$ $0=W-g\left(\frac{1}{2} t\right)$
4.0 or 3.98 (s)
ALT 2: From A to $C: W^{2}=24^{2}-2 \times 10 g$ $0=W t+\frac{1}{2}(-g) t^{2}$
4.0 or 3.98 (s)
(4)

(d)

Sketch a speed-time graph for the motion of the particle from projection to the instant that it reaches B. (No further calculations are required.)

[ 2 ]

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Covered syllabus points

Question 1

Question 2

Question 3

Question (a)

Question (b)

Question (c)

Question (d)