// Numbas version: exam_results_page_options {"name": "John's copy of Simon's copy of John's copy of Simon's copy of John's copy of Simon's copy of John's copy of Simon's copy of John's copy of Simon's copy of Particle projected up a rough inclined plane", "extensions": [], "custom_part_types": [], "resources": [["question-resources/inclined_solution_2.png", "/srv/numbas/media/question-resources/inclined_solution_2.png"]], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "question_groups": [{"pickingStrategy": "all-ordered", "questions": [{"name": "John's copy of Simon's copy of John's copy of Simon's copy of John's copy of Simon's copy of John's copy of Simon's copy of John's copy of Simon's copy of Particle projected up a rough inclined plane", "variablesTest": {"maxRuns": 100, "condition": "a<0"}, "tags": [], "parts": [{"showCorrectAnswer": true, "prompt": "

What is the normal reaction, $R$, between the particle and the plane?

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Give your answer in Newtons to 3 decimal places.

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$R = $ [[0]]

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You have not given your answer to the correct precision.

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Using the value for $R$ from part a) find the deceleration of the particle as it moves up the plane.

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Give your answer in $\\mathrm{ms^{-2}}$ to 3 decimal places.

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Find the distance that the particle will move before it comes to instantaneous rest.

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Give your answer in metres ($\\mathrm{m}$) to 3 decimal places. 

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What is my mental age?

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The acceleration of the particle along the slope.

", "definition": "-(mu*R+mass*g*sin(radians(theta)))/mass"}, "mu": {"name": "mu", "group": "Ungrouped variables", "templateType": "randrange", "description": "

The coefficient of friction between the particle and the plane.

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The mass of the particle.

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The acceleration due to gravity.

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The distance travelled before the particle comes to rest.

", "definition": "-u^2/(2*a)"}, "R": {"name": "R", "group": "Ungrouped variables", "templateType": "anything", "description": "

The normal reaction force of the plane against the particle.

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The initial speed of the particle.

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The angle of the slope.

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A particle is projected up a rough plane at a given speed. Given the angle of the slope and the coefficient friction, find the distance the particle travels before it comes to instantaneous rest.

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A particle of mass $\\var{mass} \\mathrm{kg}$ is projected with speed $\\var{u}\\ \\mathrm{ms^{-1}}$ up a rough plane. The coefficient of friction between the particle and the plane is $\\var{mu}$. The plane is inclined at an angle $\\theta = \\var{theta}^{\\circ}$ to the horizontal.

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The acceleration due to gravity is $g = 9.8 \\mathrm{ms^{-2}}$.

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A diagram can be drawn to show the forces acting on the particle.

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Here the particle is drawn in three positions, showing its original speed when it is projected from the bottom of the slope, its position at some point up the slope and its position when it instantaneously comes to rest further up the slope.

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The slope is rough so friction ($\\mu R $ N) acts down the slope, against the direction of motion.

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a)

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The normal reaction $R$, is found by resolving the forces perpendicular to the plane.

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\\begin{align}
R - mg \\cos \\theta & = 0, \\\\
R & = mg \\cos \\theta, \\\\
& = (\\var{mass} \\times 9.8) \\cos (\\var{theta}^{\\circ}), \\\\
&= \\var{precround(R,3)} \\mathrm{N}.
\\end{align}

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The normal reaction force between the particle and the plane is $\\var{precround(R,3)} \\mathrm{N}$.

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b)

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To find the acceleration of the particle we resolve the forces parallel to the plane. 

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\\begin{align}
- mg \\sin \\theta - \\mu R & = ma, \\\\
- \\var{mass} \\times 9.8 \\sin (\\var{theta}^{\\circ}) - (\\var{precround(R,3)} \\times \\var{mu}) & = \\var{mass}a, \\\\
a & = \\frac{ - \\var{mass} \\times 9.8 \\sin (\\var{theta}^{\\circ}) - (\\var{precround(R,3)} \\times \\var{mu}) }{\\var{mass}}, \\\\
& = \\var{precround(a,3)} \\mathrm{ms^{-2}}.
\\end{align}

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Therefore the deceleration of the particle is $\\var{precround(-a,3)} \\mathrm{ms^{-2}}$ as it is the negative of the acceleration.

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c)

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The particle will travel up the slope until it comes to instantaneous rest; at this point its velocity will be $0$. We can use the SUVAT equation $v^2 = u^2 + 2as$ in order to find the distance the particle will travel.

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We know that $u = \\var{u}, v= 0$ and $a = \\var{precround(a,3)}$.

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\\begin{align}
v^2 & = u^2 + 2as, \\\\
0 & = \\simplify{{u}^2+{precround(2a,3)}s}, \\\\
s & = \\frac{\\var{u}^2}{\\var{precround(-2a,3)}}, \\\\[0.5em]
& = \\var{precround(u^2/(-2*a),3)} \\mathrm{m}.
\\end{align}

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The distance the particle will travel up the plane before it comes to instantaneous rest is $\\var{precround(u^2/(-2*a),3)} \\mathrm{m}$.

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