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Solution to Test Paper 2

From Mechanics

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Solutions
1 (a) 44.1=219.8t2t=4.944.1=3 sORs=219.832=44.1Hits ground after 3 secondsAGM1

A1

A1


(M1)

(A1)

(A1)


(3 marks) M1: Use of constant acceleration

equation with v=0

A1: Correct equation

A1: Correct s


1 (b) v2=02+29.844.1v=864.36=29.4mORv=0+9.83v=29.4M1

A1

A1


(3 marks)


M1: Use of constant acceleration equation with v=0

A1: Correct equation.

A1: Correct v.


1 (c) Air resistance would slow the ball down. B1 (1 mark) B1: Sensible statement about air resistance.
(7 marks)
2 (a) Image:test2ans2.gif B1 (1 mark) B1: Correct horizontal forces.

Ignore any vertical forces.


2 (b) P=900N B1 (1 mark) B1: Correct value for P.
2 (c) P−900=20001.2P=2400+900=3300N M1

A1

A1

(1 mark) M1: Three term equation of motion

A1: Correct equation

A1: Correct P.


2 (d) 800−900=2000aa=2000−100=−0.05 ms−2Car is slowing down M1

A1

A1

A1

(4 marks) M1: Three term equation of motion

A1: Correct equation

A1: Correct a

A1: Correct statement


(9 marks)
3 (a) R=209.8=196 N M1

A1

(2 marks) M1: Use of R=mg

A1: Correct R.


3 (b) F=0.4196=78.4 N M1

A1

(2 Marks) M1: Use of \displaystyle F=\mu R

A1: Correct \displaystyle F.


3 (c) \displaystyle \begin{align}

& 100-78 \textrm{.}4=20a \\ & a=\frac{100-78 \textrm{.}4}{20}=1 \textrm{.}08 \text{ m}{{\text{s}}^{-2}} \\ \end{align}

M1

A1

A1

(3 marks) M1: Three term equation of motion

A1: Correct equation

A1: Correct \displaystyle a.


(8 marks)
4 (a) Image:test2ans4.gif B1 (1 mark) B1: Correct force diagram
4 (b) \displaystyle \begin{align}

& 100a=200-980\sin 5{}^\circ \\ & a=\frac{200-980\sin 5{}^\circ }{100}=1\textrm{.}15\ \text{m}{{\text{s}}^{-2}} \\ \end{align}

M1A1

M1

A1

(4 marks) M1: Three term equation of motion

A1: Correct equation

M1: Rearranging equation.

A1: Correct \displaystyle a.


4 (c)

\displaystyle \begin{align} & s=0\times 5+\frac{1}{2}\times 1\textrm{.}15\times {{5}^{2}} \\ & =14\textrm{.}4\ \text{m} \\ \end{align}

M1

A1

A1

(3 marks) M1: Using a constant acceleration equation

A1: Correct equation

A1: Correct distance.


(8 marks)
5 (a) \displaystyle \mathbf{v}=(6\mathbf{i}+4\mathbf{j})+(0\textrm{.}2\mathbf{i}-0\textrm{.}4\mathbf{j})t M1

A1

(2 marks) M1: Use of \displaystyle \mathbf{v}=\mathbf{u}+\mathbf{a}t

A1: Correct expression


5 (b) \displaystyle \begin{align}

& 4-0\textrm{.}4t=0 \\ & t=10 \ \text{s}\\ \end{align}

M1

A1

A1

(3 marks) M1: Using \displaystyle \mathbf{j} component equal to zero.

A1: Correct equation.

A1: Correct time.

5 (c) \displaystyle \begin{align}

& \mathbf{r}=(6\mathbf{i}+4\mathbf{j})\times 30+\frac{1}{2}(0\textrm{.}2\mathbf{i}-0\textrm{.}4\mathbf{j})\times {{30}^{2}} \\ & =270\mathbf{i}-60\mathbf{j} \\ & r=\sqrt{{{270}^{2}}+{{60}^{2}}}=277\ \text{m} \\ \end{align}

M1

A1

M1

A1

(4 marks) M1: Using \displaystyle \mathbf{r}=\mathbf{u}t+\frac{1}{2}\mathbf{a}{{t}^{2}}

A1: Correct position vector.

M1: Finding distance.

A1: Correct distance

(8 marks)


KEY M1: Method Mark

A1: Accuracy Mark following a method mark

B1: Accuracy Mark not following a method mark

AG: Answer Given in Question – Working must justify answer.

TOTAL: 40 Marks

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