Pearson Edexcel IAL Mathematics Unit M1: Mechanics 1 Question Bank

$1.4=\frac{1}{2} a \times 2^{2}$

M1

$a=0.7\left(\mathrm{~m} \mathrm{~s}^{-2}\right)^{*}$ GIVEN ANSWER

A1*

(2)

Inextensibility of string

B1

(1)

$$3g-T=3\times0.7$$T=3g-2.1$$\text{Resultant}=2T\cos45^{\circ}$$

OR

$$\text{Resultant}=\sqrt{T^2+T^2}$$

OR

$$\text{Resultant}=\frac{T}{\cos45^{\circ}}$$$$\text{Resultant}=39\text{ N}$$

Accept $38.6\text{ N}$.

M1 A1 M1 A1

Marking guidance for this part:
M1 for an equation of motion for B with usual rules.
A1 for the correct equation.
M1 for a correct expression in terms of T.
A1 for 39 or 38.6 N.

$T-F=4 \times 0.7 \quad$ (for $A) \quad$ OR $\quad 3 g-F=7 \times 0.7 \quad$ (whole system)

M1 A1

$R=4 g ; F=\mu \times R$

B1; B1

$27.3-\mu \times 4 g=4 \times 0.7 \quad$ OR $\quad 3 g-\mu \times 4 g=7 \times 0.7$

DM1

$\mu=0.625$ or 0.63

A1

(6)

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

$T=m a \quad($ allow -a)

B1

(1)

$$4 m g \sin \alpha-T-F=4 m a \quad \text { OR } \quad 4 m g \sin \alpha-F=5 m a$$

(allow -a ) (allow -a )

M1A1

$F=\frac{1}{4} R$

B1

$R=4 m g \cos \alpha$

M1A1

Solve for T in terms of m g only

DM1

$T=\frac{8 m g}{25}$ oe

A1

(7)

$2 T \sin \frac{1}{2} \alpha$ oe e.g. $\sqrt{T^{2}+T^{2}-2 T^{2} \cos \alpha}$ using cos rule

Or $\frac{T \sin \alpha}{\sin \left(90^{\circ}-\frac{1}{2} \alpha\right)}$ using sine rule

Or $\sqrt{(T-T \cos \alpha)^{2}+(T \sin \alpha)^{2}}$ using components and Pythag.

M1 A1

Substitute for T and trig

M1

$\frac{8 m g \sqrt{10}}{125}$ oe, 2 m or 2.0 m or 1.98 m or 0.2 m g or better

A1

(4)

e.g. Tension will be the same throughout a section of the string.

B1 (1)

(13)

Notes for question 7

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

B1

Seen or implied

Correct use of F=m a

M1

Must have added the forces

Acceleration
$$=(\mathbf{i}+2.5 \mathbf{j})\left(\mathrm{ms}^{-2}\right)$$

A1

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
$$=((2+b) \mathbf{i}+(5+c) \mathbf{j})(\mathrm{N})$$

B1

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

M1

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