Edexcel A-Level Mathematics AS Unit M1 Mechanics 1 Questions

$\mathrm{M}(D), m g \times 1.2=30 g \times 0.8$
Other possible equations:

( ↑ ) R=m g+30 g

$\mathrm{M}(A) 2.5 m g+30 g \times 4.5=3.7 R$$\mathrm{M}(G) 30 g \times 2=1.2 R$

M( C $) \quad m g \times 2=0.8 R$$\mathrm{M}(B) 2.5 m g+30 g \times 0.5=1.3 R$$m=20(\mathrm{~kg})$
N.B. Allow an inequality if they state $m=20(\mathrm{~kg})$ at the end

(3)

$$\mathrm{M}(D), X g \times 3.7+20 g \times 1.2=30 g \times 1.3$$

N.B. Allow inequality $\geq \ldots$ the correct way round for M1A1ft

M1A1ft

Other possible equations:

( ↑ ) S=m g+30 g+X g

$\mathrm{M}(A) \quad 2.5 m g+30 g \times 5=3.7 S$

where m is their answer from (a).

$\mathrm{M}(G) \quad 30 g \times 2.5=1.2 S+X g \times 2.5$$\mathrm{M}(B) 2.5 m g+X g \times 5=1.3 S$$X=\frac{150}{37}, 4.1$ or better $(4.05405 \ldots)$

A1

(3)

The mass of the block is concentrated at a point. oe
N.B. Must mention either mass or weight and 'acting at a point' or 'concentrated at a point'.

(1)

(7)

Notes for question 3

$(5 \mathbf{i}-8 \mathbf{j})+5(-\lambda \mathbf{i}+2 \lambda \mathbf{j})\left(\mathrm{m} \mathrm{s}^{-1}\right)$ isw
(2)

$13=\sqrt{(5-5 \lambda)^{2}+(-8+10 \lambda)^{2}}$

M1 A1

$169=25-50 \lambda+25 \lambda^{2}+64-160 \lambda+100 \lambda^{2}$$25 \lambda^{2}-42 \lambda-16=0^{*}$

A1* cso

(3)

( -2 i+4 j ) seen or implied
(5 i-8 j)+(-2 i+4 j) 4

$339^{\circ}$
(5)

(10)

Notes for question 4

(a)
Or column vector

Complete method to find the velocity when t=4.

Correct unsimplified expression. Note the correct velocity is v=-3 i+8 j

Use their velocity vector at t=4 with trig to find a relevant angle.

Cao. Degrees sign not required.

N.B. if they work with both values of $\lambda$, can score max all the marks except the last one.

QUESTION
NUMBER

SCHEME

MARKS

5

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)

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)

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)

Speed-time graph:
B1 shape
B1ft labels
(2)

Resultant force
$$=(-2 \mathbf{i}+3 \mathbf{j}+4 \mathbf{i}+2 \mathbf{j})=(2 \mathbf{i}+5 \mathbf{j})(\mathrm{N})$$
Seen or implied

Correct use of F=m a
Must have added the forces

Acceleration
$$=(\mathbf{i}+2.5 \mathbf{j})\left(\mathrm{ms}^{-2}\right)$$
Isw if they find the
magnitude

[3]

$$\mathbf{v}=(3\mathbf{i}-4\mathbf{j})+3\mathbf{a}$$
Using $\mathbf{a}=\mathbf{i}+2.5\mathbf{j}$,
$$\mathbf{v}=6\mathbf{i}+3.5\mathbf{j}$$$$|\mathbf{v}|=\sqrt{6^2+3.5^2}=6.95\text{ m s}^{-1}$$
M1 A1ft dM1 A1

Resultant force
$$=((2+b) \mathbf{i}+(5+c) \mathbf{j})(\mathrm{N})$$
i's andj's must be collected
but can be unsimplified,
seen or implied.

Correct use of direction to form equation in b and c only
$$\text { e.g } 2+b=5+c$$

or $2+b=\lambda$ and $5+c=\lambda$ with $\lambda$ eliminated
MO if they use a specific
numerical value for $\lambda$ or
they use 2+b=-(5+c)

$b-c=3^{*}$

A1*

Obtain given answer from
correct working

[3]

Use the magnitude of the resultant to form an equation in b and c:
$$200=(2+b)^2+(5+c)^2$$
The two possible forces are
$$\mathbf{F}_3=(8\mathbf{i}+5\mathbf{j})\text{ N}$$$$\mathbf{F}_3=(-12\mathbf{i}-15\mathbf{j})\text{ N}$$
M1 A1ft dM1 A1