Solution to Test Paper 2
From Mechanics
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|+ Solutions | |+ Solutions | ||
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| 1 (a) | | 1 (a) | ||
| <math>\begin{align} | | <math>\begin{align} | ||
| - | & 44 | + | & 44\textrm{.}1=\frac{1}{2}\times 9\textrm{.}8{{t}^{2}} \\ |
| - | & t=\sqrt{\frac{44 | + | & t=\sqrt{\frac{44\textrm{.}1}{4\textrm{.}9}}=3\text{ s} \\ |
| - | \ | + | & \\ |
| - | + | & \text{OR} \\ | |
| - | + | & \\ | |
| - | + | & s=\frac{1}{2}\times 9\textrm{.}8\times {{3}^{2}}=44\textrm{.}1 \\ | |
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| - | & s=\frac{1}{2}\times 9 | + | |
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& \therefore \text{Hits ground after 3 seconds} \\ | & \therefore \text{Hits ground after 3 seconds} \\ | ||
\end{align}</math> | \end{align}</math> | ||
| - | + | AG | |
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| 1 (b) | | 1 (b) | ||
| <math>\begin{align} | | <math>\begin{align} | ||
| - | & {{v}^{2}}={{0}^{2}}+2\times 9 | + | & {{v}^{2}}={{0}^{2}}+2\times 9\textrm{.}8\times 44\textrm{.}1 \\ |
| - | & v=\sqrt{864 \textrm{.}36}=29 | + | & v=\sqrt{864\textrm{.}36}=29\textrm{.}4m \\ |
| + | & \\ | ||
| + | & \text{OR} \\ | ||
| + | & \\ | ||
| + | & v=0+9\textrm{.}8\times 3 \\ | ||
| + | & v=29\textrm{.}4 \\ | ||
\end{align}</math> | \end{align}</math> | ||
| - | OR | ||
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| - | <math>\begin{align} | ||
| - | & v=0+9 \textrm{.}8\times 3 \\ | ||
| - | & v=29 \textrm{.}4 \text{ m}{{\text{s}}^{-1}} \\ | ||
| - | \end{align}</math> | ||
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& 800-900=2000a \\ | & 800-900=2000a \\ | ||
& a=\frac{-100}{2000}=-0\textrm{.}05 \text{ m}{{\text{s}}^{-2}} \\ | & a=\frac{-100}{2000}=-0\textrm{.}05 \text{ m}{{\text{s}}^{-2}} \\ | ||
| + | & \text{Car is slowing down} \\ | ||
\end{align}</math> | \end{align}</math> | ||
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| '''(8 marks)''' | | '''(8 marks)''' | ||
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Current revision
| 1 (a) |  9.8t2t= 4.944.1=3 sORs=219.832=44.1 Hits ground after 3 secondsAG | M1
A1 A1
(A1) (A1)
| (3 marks) | M1: Use of constant acceleration
equation with A1: Correct equation A1: Correct
|
| 1 (b) | 9.844.1v= 864.36=29.4mORv=0+9.83v=29.4 | M1
A1 A1
| (3 marks)
| M1: Use of constant acceleration equation with A1: Correct equation. A1: Correct
|
| 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) | | B1 | (1 mark) | B1: Correct value for |
| 2 (c) | 1.2P=2400+900=3300N | M1
A1 A1 | (1 mark) | M1: Three term equation of motion
A1: Correct equation A1: Correct
|
| 2 (d) | | M1
A1 A1 A1 | (4 marks) | M1: Three term equation of motion
A1: Correct equation A1: Correct A1: Correct statement
|
| (9 marks) | ||||
| 3 (a) | 9.8=196 | M1
A1 | (2 marks) | M1: Use of A1: Correct
|
| 3 (b) | 196=78.4 | M1
A1 | (2 Marks) | M1: Use of  RA1: Correct
|
| 3 (c) | | M1
A1 A1 | (3 marks) | M1: Three term equation of motion
A1: Correct equation A1: Correct
|
| (8 marks) | ||||
| 4 (a) | 
| B1 | (1 mark) | B1: Correct force diagram |
| 4 (b) |  a=100200−980sin5=1.15 ms−2 | M1A1
M1 A1 | (4 marks) | M1: Three term equation of motion
A1: Correct equation M1: Rearranging equation. A1: Correct
|
| 4 (c) |
| M1
A1 A1 | (3 marks) | M1: Using a constant acceleration equation
A1: Correct equation A1: Correct distance.
|
| (8 marks) | ||||
| 5 (a) | | M1
A1 | (2 marks) | M1: Use of A1: Correct expression
|
| 5 (b) | | M1
A1 A1 | (3 marks) | M1: Using A1: Correct equation. A1: Correct time. |
| 5 (c) | 30+21(0.2i−0.4j)302=270i−60jr= 2702+602=277 m | M1
A1 M1 A1 | (4 marks) | M1: Using 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
