From Mechanics
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- | | 1 | + | | |
- | | 2 | + | | |
- | | 3 | + | | |
- | | 4 | + | | '''(8 marks)''' |
- | | 5 | + | | |
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| |- | | |- |
| | 5 (a) | | | 5 (a) |
- | | 2 | + | | <math>\mathbf{v}=(6\mathbf{i}+4\mathbf{j})+(0\textrm{.}2\mathbf{i}-0\textrm{.}4\mathbf{j})t</math> |
- | | 3 | + | | M1 |
- | | 4 | + | |
- | | 5 | + | A1 |
| + | | (2 marks) |
| + | | M1: Use of <math>\mathbf{v}=\mathbf{u}+\mathbf{a}t</math> |
| + | |
| + | A1: Correct expression |
| + | |
| | | |
| |- | | |- |
| | 5 (b) | | | 5 (b) |
- | | 2 | + | | <math>\begin{align} |
- | | 3 | + | & 4-0\textrm{.}4t=0 \\ |
- | | 4 | + | & t=10 \ \text{s}\\ |
- | | 5 | + | \end{align}</math> |
| + | |
| + | | M1 |
| + | |
| + | A1 |
| + | |
| + | A1 |
| + | |
| + | | (3 marks) |
| + | | M1: Using <math>\mathbf{j}</math> component equal to zero. |
| + | |
| + | A1: Correct equation. |
| + | |
| + | A1: Correct time. |
| + | |
| |- | | |- |
| | 5 (c) | | | 5 (c) |
- | | 2 | + | | <math>\begin{align} |
- | | 3 | + | & \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}} \\ |
- | | 4 | + | & =270\mathbf{i}-60\mathbf{j} \\ |
- | | 5 | + | & r=\sqrt{{{270}^{2}}+{{60}^{2}}}=277\ \text{m} \\ |
| + | \end{align}</math> |
| + | |
| + | | M1 |
| + | |
| + | A1 |
| + | |
| + | M1 |
| + | |
| + | A1 |
| + | |
| + | | (4 marks) |
| + | | M1: Using <math>\mathbf{r}=\mathbf{u}t+\frac{1}{2}\mathbf{a}{{t}^{2}}</math> |
| + | |
| + | 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''' |
Revision as of 16:43, 19 January 2011
Solutions
1 (a)
| 44 .1=21 9.8t2t= 4.944.1=3 sORs=21 9.8 32=44.1AG Hits ground after 3 seconds | M1
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+2 9.8 44.1v= 864.36=29 .4 ms−1ORv=0+9.8 3v=29.4 ms−1 | M1
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)
|
| 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=2000 1.2P=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−2Carisslowingdown | 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=20 9.8=196 N
| M1
A1
| (2 marks)
| M1: Use of R=mg
A1: Correct R.
|
3 (b)
| F=0.4 196=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)
|
| 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