// Numbas version: exam_results_page_options {"name": "Dynamics - Forces ", "duration": 0, "metadata": {"description": "

To be used on the Mechanics wiki page under Dynamics - Forces.

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a)

The weight of a particle in Newtons is given by

\\begin{align} W & = mg \\\\
& = \\var{m} \\times 9.8 \\\\
& = \\var{m*9.8}, \\end{align}

where $g$ is the force due to gravity acting on an object. The particle weighs $\\var{precround(m*9.8,3)} \\, \\mathrm{N}$.

b)

We can use the relationship $F=ma$ to find the acceleration $a$.

\\begin{align} F & = ma \\\\
\\var{F} & = \\var{m} \\times a \\\\
a & = \\var{precround(F/m,3)}. \\end{align}
The acceleration is $\\var{precround(F/m,3)} \\, \\mathrm{ms}^{-2}$.

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Find the weight in Newtons of this particle.

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Find the acceleration in $\\mathrm{ms^{-2}}$ of the particle when it is acted on by a resultant force of $\\var{F}\\ \\mathrm{N}$. Give your answer to 3 decimal places.

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Consider a particle of mass $\\var{m} \\, \\mathrm{kg}$. The acceleration due to gravity is $9.8 \\, \\mathrm{ms^{-2}}$.

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particles mass

"}, "F": {"definition": "random(1..15#1)", "templateType": "randrange", "group": "Ungrouped variables", "name": "F", "description": "

Force

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Find the weight of a particle when the mass and acceleration due to gravity is known. Using $F=ma$ to find the acceleration of the particle when it is acted on by a resultant force.

", "licence": "Creative Commons Attribution 4.0 International"}, "type": "question", "showQuestionGroupNames": false, "question_groups": [{"name": "", "pickingStrategy": "all-ordered", "pickQuestions": 0, "questions": []}]}, {"name": "Particle in vertical equilibrium, accelerating horizontally", "extensions": [], "custom_part_types": [], "resources": [["question-resources/Force_image.png", "/srv/numbas/media/question-resources/Force_image.png"], ["question-resources/improvement1.png", "/srv/numbas/media/question-resources/improvement1.png"]], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "contributors": [{"name": "Christian Lawson-Perfect", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/7/"}, {"name": "Amy Chadwick", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/505/"}], "functions": {}, "ungrouped_variables": ["a", "K", "FA", "FD", "a2", "K2", "FA2", "downforce", "FD2"], "tags": [], "advice": "

a)

To find the resultant force in a certain direction when there is more than one force acting on an object you can resolve the forces. You resolve in the direction of acceleration. Using the letter $R$, with an arrow will indicate the direction the forces are being resolved in, for example $R(\\rightarrow)$.

In this case $R(\\rightarrow)$ means apply $F=ma$ in the positive direction. So we need to resolve the equation

\\begin{align}F & = ma \\\\
C - \\var{FA} & = \\var{K} \\times \\var{a} \\\\
C & = \\var{K*a} + \\var{FA} \\\\
& = \\var{K*a + FA}. \\end{align}

Here the forces, $F=C - \\var{FA}$ because $\\var{FA}$ is acting in the opposite direction to $C$. The mass is $m=\\var{K}$ and the acceleration is $a=\\var{a}$. Therefore $C = \\var{K*a+FA}\\ \\mathrm{N}$.

b)

In this case we resolve in the direction perpendicular to acceleration so we use $R(\\uparrow)$ which means apply $F=ma$ in the vertical direction, where acceleration, $a=0$ because there is no vertical acceration.

\\begin{align} F & = ma \\\\
B - \\var{FD}g & = \\var{K} \\times 0 \\\\
B & = 0 + \\left(\\var{FD} \\times 9.8\\right) \\\\
& = \\var{FD*9.8} \\end{align}
Here the forces, $F=B - \\var{FD}g$ because $\\var{FD}g$ is acting downwards which is the opposite direction to $B$. Therefore $B = \\var{FD*9.8} \\ \\mathrm{N}$.

c)

It is easier to take the positive direction as the direction of the acceleration. Therefore we use $R(\\leftarrow)$ because we are decelerating.

\\begin{align} F & = ma \\\\
\\var{FA2} - C & = \\var{K2} \\times \\var{a2} \\\\
C & = \\var{FA2} - \\left( \\var{K2} \\times \\var{a2} \\right) \\\\
& = \\var{FA2 - K2*a2}. \\end{align}

Here the forces, $F = \\var{FA2} - C$ because $\\var{FA2}$ is in the direction of acceleration and $C$ is acting in the opposite direction. The mass is $m = \\var{K2}$ and $a=\\var{a2}$. Therefore $C = \\var{FA2 - K2*a2} \\ \\mathrm{N}$.

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Suppose that the acceleration $a = \\var{a}$ and the mass of the particle $K = \\var{K} \\ \\mathrm{kg}$. Find the force $C \\ \\mathrm{N}$ if $A = \\var{FA}\\ \\mathrm{N}$.

$C = $ [[0]] $\\mathrm{N}$

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You have not given your answer to the correct precision.

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Find the magnitude of the force $B$ if $D =\\simplify{{FD}g} \\ \\mathrm{N}$.

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Consider another particle which has mass $\\var{K2} \\ \\mathrm{kg}$ and is now decelerating horizontally at $\\var{a2}\\ \\mathrm{ms^{-2}}$. If $A = \\var{FA2}\\ \\mathrm{N}$ what is $C$?

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Consider the following diagram of a particle of mass $K \\ \\mathrm{kg}$, moving left-to-right. The forces $A, B, C$ and $D$ are acting upon it, producing a horizontal acceleration of $a \\, \\mathrm{ms^{-2}}$

The acceleration due to gravity is $9.8 \\, \\mathrm{ms}^{-2}$. Answer all the following questions to 3 decimal places.

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acceleration

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mass 2

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mass

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Force A

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downforce D 2

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Force D

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downforce for part d)

"}, "a2": {"definition": "random(0.25..5#0.25)", "templateType": "randrange", "group": "Ungrouped variables", "name": "a2", "description": "

acceleration

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Using $F=ma$ to find magnitudes of unknown forces acting on an accelerating body.

a) Use $W=mg = \\var{mass1} \\times \\var{g}$. So the weight is $\\var{mass1*g} \\ \\mathrm{N}$.

b) To find the mass we use $W=mg$ rearranged for $m$. This gives $m = \\frac{W}{g} = \\frac{\\var{weight1}}{\\var{g}}$. So the mass is $\\var{precround(weight1/g,3)} \\ \\mathrm{kg}$.

c) Use $F=ma$ with $m=\\var{mass2}$ and $a=\\var{a1}$. This gives $F=\\var{mass2} \\times \\var{a1} = \\var{mass2*a1}$. So the force required is $\\var{mass2*a1} \\ \\mathrm{N}$.

d) Use $F=ma$ rearranged for $a$. This gives $a = \\frac{F}{m} = \\frac{\\var{force1}}{\\var{mass3}}$. So the acceleration is $a=\\var{precround(force1/mass3,3)} \\ \\mathrm{ms^{-2}}$.

e) Use $F=ma$ rearranged for $m$. This gives $m = \\frac{F}{a} = \\frac{\\var{force2}}{\\var{decelerate}}$. So the mass is $m=\\var{precround(force2/decelerate,3)} \\ \\mathrm{kg}$.

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Find the weight in Newtons of a box that has mass $\\var{mass1} \\, \\mathrm{kg}$.

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If the weight of a particle is $\\var{weight1} \\, \\mathrm{N}$ what is its mass?

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Find the force required to accelerate a $\\var{mass2} \\, \\mathrm{kg}$ mass at a rate of $\\var{a1} \\, \\mathrm{ms^{-2}}$.

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Find the acceleration when a particle of mass $\\var{mass3} \\, \\mathrm{kg}$ is acted on by a resultant force of $\\var{force1} \\, \\mathrm{N}$.

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A moving object is acted upon by a force of $\\var{force2} \\, \\mathrm{N}$ which causes it to decelerate at a rate of $\\var{decelerate} \\, \\mathrm{ms^{-2}}$. Find the mass of the object.

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Using $g= 9.8\\ \\mathrm{ms^{-2}}$ answer the following questions to 3 decimal places.

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gravity

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acceleration

"}, "force1": {"definition": "random(80..300#1)", "templateType": "randrange", "group": "Ungrouped variables", "name": "force1", "description": "

force acting on particle

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Solving questions using $F=ma$ and $W=mg$.

a)

We resolve $F=ma$ in the horizontal direction of acceleration.

\\begin{align}
F & = ma \\\\
\\var{force1} - \\var{friction} & = \\var{mass} \\times a \\\\
a & = \\frac{\\var{force1} - \\var{friction}}{\\var{mass}} \\\\
& = \\var{precround((force1 - friction)/mass,3)} \\, \\mathrm{ms^{-2}}
\\end{align}

Therefore the acceleration of the particle is $\\var{a}\\, \\mathrm{ms^{-2}}$.

b)

To find the distance travelled we can use the equation $s = ut+\\frac{1}{2}at^2$. The particle begins at rest, so the initial velocity $u=0$, and we have the acceleration $a=\\var{a}$ from part a).

\\begin{align}
s & = ut+ \\frac{1}{2}at^2 \\\\
& = \\left(0 \\times t \\right) + \\left( \\frac{1}{2} \\times \\var{a} \\times \\var{time}^2 \\right) \\\\
& = \\var{precround(0.5*a*time^2,3)} \\, \\mathrm{m}
\\end{align}

Therefore the distance travelled is $\\var{precround(0.5*a*time^2,3)}\\ \\mathrm{m}$.

c)

We resolve $F=ma$ in the vertical direction, where acceleration is zero. The normal reaction, $R$, is the force which acts perpendicular to the surface, therefore $R$ is in the upwards direction. So $R$ is acting in the opposite direction to the weight of the particle which is $W=mg$.

\\begin{align}
F & = ma \\\\
R - W &= m \\times 0 \\\\
R - mg & = 0 \\\\
R & = mg \\\\
& = \\var{mass} \\times 9.8 \\\\
& = \\var{precround(mass*9.8,3)} \\, \\mathrm{N}
\\end{align}

So the normal reaction force has magnitude $\\var{precround(mass*9.8,3)}\\, \\mathrm{N}$.

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What is the acceleration of the particle, in $\\mathrm{ms^{-2}}$?

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In metres, how far does the particle travel in the first $\\var{time}$ seconds?

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Find the magnitude in $\\mathrm{N}$ of the normal reaction $R$ between the particle and the surface when $g= 9.8 \\, \\mathrm{ms^{-2}}$.

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A particle of mass $\\var{mass}\\, \\mathrm{kg}$, which is initially at rest, is pushed across a rough surface by a horizontal force of $\\var{force1} \\, \\mathrm{N}$ against a frictional force of $\\var{friction}\\, \\mathrm{N}$.

Give your answers to the following questions to 3 decimal places.

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Find the acceleration, distance moved and magnitude of the normal reaction of a particle being pulled along a rough horizontal plane.

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First we can draw a diagram of the forces acting on the block.

Remember that when resolving $F=ma$ we have to work in $\\mathrm{kg}$ for mass. So we need to convert $\\var{showgrams}\\ \\mathrm{g}$ to $\\var{mass} \\ \\mathrm{kg}$.

a)

When the block is moving upwards we resolve $F=ma$ in the vertical direction, with upwards being positive.

\\begin{align}
F & = ma \\\\
T - mg & = ma \\\\
T & = \\left(\\var{mass} \\times \\var{g}\\right) + \\left( \\var{mass} \\times \\var{a1} \\right) \\\\
& = \\var{precround(mass*g + mass*a1,3)}.
\\end{align}

The tension in the rope is $\\var{precround(mass*g + mass*a1,3)} \\ \\mathrm{N}$.

b)

When the block is moving downwards we resolve vertically downwards, as this is the direction of acceleration.

\\begin{align}
F & = ma \\\\
mg - T & = ma \\\\
T & = \\left( \\var{mass} \\times \\var{g} \\right) - \\left( \\var{mass} \\times \\var{b}\\right) \\\\
& = \\var{precround(mass*g - mass*b,3)}.
\\end{align}

The tension in the rope is $\\var{precround(mass*g - mass*b,3)} \\ \\mathrm{N}$.

c)

Here, constant speed means acceleration $a=0$. Therefore we can resolve $F=ma$ in the upward direction of acceleration but set $a=0 \\ \\mathrm{ms^{-2}}$.

\\begin{align}
F & = ma \\\\
T - mg & = m \\times 0 \\\\
T & = m \\times g \\\\
& = \\var{mass} \\times 9.8 \\\\
& = \\var{precround(mass*9.8,3)}
\\end{align}

The tension in the rope is $\\var{precround(mass*g,3)} \\ \\mathrm{N}$.

d)

The block is moving downwards but decelerating at a rate of $a= -\\var{a3} \\ \\mathrm{ms^{-2}}$.

\\begin{align}
F & = ma \\\\
mg - T & = ma \\\\
T & = mg - ma \\\\
& = \\left( \\var{mass} \\times \\var{g} \\right) - \\left(\\var{mass} \\times \\var{-a3}\\right) \\\\
& = \\var{precround(mass*g + mass*a3,3)}
\\end{align}

The tension in the rope is $\\var{precround(mass*g+mass*a3,3)} \\ \\mathrm{N}$.

", "rulesets": {}, "parts": [{"precisionType": "dp", "prompt": "

When the block moves upward with an acceleration of $\\var{a1}\\ \\mathrm{ms^{-2}}$.

", "precisionMessage": "You have not given your answer to the correct precision.", "allowFractions": false, "variableReplacements": [], "maxValue": "mass*g+(mass*a1)", "strictPrecision": false, "minValue": "mass*g+(mass*a1)", "variableReplacementStrategy": "originalfirst", "precisionPartialCredit": 0, "correctAnswerFraction": false, "showCorrectAnswer": true, "precision": "3", "scripts": {}, "marks": 1, "type": "numberentry", "showPrecisionHint": false}, {"precisionType": "dp", "prompt": "

When the block moves downwards with an acceleration of $\\var{a1+a2}\\ \\mathrm{ms^{-2}}$.

", "precisionMessage": "You have not given your answer to the correct precision.", "allowFractions": false, "variableReplacements": [], "maxValue": "mass*g-mass*(a1+a2)", "strictPrecision": false, "minValue": "mass*g-mass*(a1+a2)", "variableReplacementStrategy": "originalfirst", "precisionPartialCredit": 0, "correctAnswerFraction": false, "showCorrectAnswer": true, "precision": "3", "scripts": {}, "marks": 1, "type": "numberentry", "showPrecisionHint": false}, {"precisionType": "dp", "prompt": "

When the block of mass moves upward with a constant speed of $\\var{u}ms^{-1}$.

", "precisionMessage": "You have not given your answer to the correct precision.", "allowFractions": false, "variableReplacements": [], "maxValue": "mass*g", "strictPrecision": false, "minValue": "mass*g", "variableReplacementStrategy": "originalfirst", "precisionPartialCredit": 0, "correctAnswerFraction": false, "showCorrectAnswer": true, "precision": "3", "scripts": {}, "marks": 1, "type": "numberentry", "showPrecisionHint": false}, {"precisionType": "dp", "prompt": "

When the block of mass moves downward with a deceleration of $\\var{a3}ms^{-2}$.

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A block of mass $\\var{showgrams} \\, \\mathrm{g}$ is attached to a vertical rope.

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

Find the tension in the rope, in Newtons to 3 decimal places, in the following situations.

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LaTeX encoding of the weight in grams, with thousands separator.

"}, "mass": {"definition": "random(1..10#0.25)", "templateType": "randrange", "group": "Ungrouped variables", "name": "mass", "description": ""}}, "metadata": {"description": "

A mass attached to a vertical rope. Finding the tension in the rope for different accelerations. Using $F=ma$.

a)

We are told that the caravan is decelerating at a rate of $a= \\var{a} \\ \\mathrm{ms^{-2}}$. We know $u=\\var{u}$ and $s=\\var{s}$ and we want $v$, therefore we can use the equation $v^2=u^2+2as$.

\\begin{align}
v^2 & = u^2 + 2as \\\\
& = \\var{u}^2 + \\left( 2 \\times \\var{a} \\times \\var{s} \\right) \\\\
& = \\var{u^2 + 2*a*s}
\\end{align}

Therefore $v = \\sqrt{\\var{u^2 + 2*a*s}} = \\var{v}$. So the speed is $\\var{v} \\ \\mathrm{ms^{-1}}$.

b)

Having worked out the speed $v$ of the caravan after it has travelled $\\var{s} \\, \\mathrm{m}$, we can use the formula $v = u + at$ to find the time taken.

\\begin{align}
v &= u + at \\\\
t &= \\frac{v-u}{a} \\\\
&= \\simplify[!basic]{({v}-{u})/{a}} \\\\
&= \\var{precround((v-u)/a,3)}
\\end{align}

The caravan takes $\\var{precround((v-u)/a,3)} \\, \\mathrm{s}$ to travel $\\var{s} \\, \\mathrm{m}$.

c)

To find the tension, $T$ in Newtons, we resolve in the direction of acceleration, where $a=\\var{a}$ and $T$ is acting in the opposite direction.

\\begin{align}
F & = ma \\\\
- T & = \\var{mass} \\times \\var{a} \\\\
T & = \\var{precround(mass*-a,3)}.
\\end{align}

The tension in the rope is $\\var{precround(mass*-a,3)} \\ \\mathrm{N}$.

d)

If the caravan experiences a resistance to motion of magnitude $\\var{resistance}N$ this resistance will act in the opposite direction to acceleration.

\\begin{align} F & = ma \\\\
- T - \\var{resistance} & = \\var{mass} \\times \\var{a}\\\\
T & = \\left(\\var{mass} \\times \\var{-a}\\right) - \\var{resistance} \\\\
& = \\var{precround(mass*-a - resistance,3)}. \\end{align}

The tension in the rope is $\\var{precround(mass*-a-resistance,3)} \\ \\mathrm{N}$.

", "rulesets": {}, "parts": [{"precisionType": "dp", "prompt": "

Find the speed of the caravan in $\\mathrm{ms}^{-1}$ after it has travelled $\\var{s} \\, \\mathrm{m}$.

", "precisionMessage": "You have not given your answer to the correct precision.", "allowFractions": false, "variableReplacements": [], "precision": "3", "maxValue": "(u^2+2*a*s)^(1/2)", "minValue": "(u^2+2*a*s)^(1/2)", "variableReplacementStrategy": "originalfirst", "strictPrecision": false, "correctAnswerFraction": false, "showCorrectAnswer": true, "precisionPartialCredit": 0, "scripts": {}, "marks": 1, "type": "numberentry", "showPrecisionHint": false}, {"precisionType": "dp", "prompt": "

How many seconds does the caravan take to travel $\\var{s} \\, \\mathrm{m}$?

", "precisionMessage": "You have not given your answer to the correct precision.", "allowFractions": false, "variableReplacements": [{"variable": "v", "part": "p0", "must_go_first": false}], "precision": "3", "maxValue": "(v-u)/a", "minValue": "(v-u)/a", "variableReplacementStrategy": "originalfirst", "strictPrecision": false, "correctAnswerFraction": false, "showCorrectAnswer": true, "precisionPartialCredit": 0, "scripts": {}, "marks": 1, "type": "numberentry", "showPrecisionHint": false}, {"precisionType": "dp", "prompt": "

Find the tension in Newtons in the rope which pulls the caravan.

", "precisionMessage": "You have not given your answer to the correct precision.", "allowFractions": false, "variableReplacements": [], "precision": "3", "maxValue": "-(mass*a)", "minValue": "-(mass*a)", "variableReplacementStrategy": "originalfirst", "strictPrecision": false, "correctAnswerFraction": false, "showCorrectAnswer": true, "precisionPartialCredit": 0, "scripts": {}, "marks": 1, "type": "numberentry", "showPrecisionHint": false}, {"precisionType": "dp", "prompt": "

If the caravan does experience a resistance to motion of magnitude $\\var{resistance} \\, \\mathrm{N}$, what would the tension in the rope be, in Newtons?

", "precisionMessage": "You have not given your answer to the correct precision.", "allowFractions": false, "variableReplacements": [], "precision": "3", "maxValue": "-mass*a - resistance", "minValue": "-mass*a - resistance", "variableReplacementStrategy": "originalfirst", "strictPrecision": false, "correctAnswerFraction": false, "showCorrectAnswer": true, "precisionPartialCredit": 0, "scripts": {}, "marks": 1, "type": "numberentry", "showPrecisionHint": false}], "statement": "

A caravan of mass $\\var{mass} \\ \\mathrm{kg}$ is pulled on a rope by a car along a straight horizontal road. It decelerates at a constant rate of $\\var{-a} \\ \\mathrm{ms^{-2}}$ from an initial speed of $\\var{u} \\ \\mathrm{ms^{-1}}$. There is no resistance to motion.

Give your answers to each of the following questions to 3 decimal places.

", "variable_groups": [], "variablesTest": {"maxRuns": 100, "condition": ""}, "preamble": {"css": "", "js": ""}, "variables": {"a": {"definition": "random(-5..-0.5#0.25)", "templateType": "randrange", "group": "Ungrouped variables", "name": "a", "description": "

negative because deceleration

"}, "resistance": {"definition": "random(10..30#1)", "templateType": "randrange", "group": "Ungrouped variables", "name": "resistance", "description": ""}, "s": {"definition": "random(10..30#1)", "templateType": "randrange", "group": "Ungrouped variables", "name": "s", "description": ""}, "mass": {"definition": "random(50..400#1)", "templateType": "randrange", "group": "Ungrouped variables", "name": "mass", "description": ""}, "v": {"definition": "precround((u^2+2*a*s)^(1/2),3)", "templateType": "anything", "group": "Ungrouped variables", "name": "v", "description": "

3dp

"}, "u": {"definition": "random(25..30#0.5)", "templateType": "randrange", "group": "Ungrouped variables", "name": "u", "description": "

initial speed

"}}, "metadata": {"description": "

A caravan pulled along by a car. Question uses SUVAT equations and $F=ma$.

a)

We resolve $F=ma$ in the upwards positive direction as acceleration is acting vertically upwards, the normal reaction, $R \\ \\mathrm{N}$, will act in the opposite direction to the weight of the ball.

\\begin{align}
F & = ma \\\\
R - mg &= ma \\\\
R & = mg + ma \\\\
& = \\left(\\var{mass}\\times \\var{g}\\right) + \\left(\\var{mass} \\times \\var{a}\\right) \\\\
& = \\var{precround(mass*g + mass*a,3)}.
\\end{align}

The normal reaction on the ball has magnitude $\\var{precround(mass*g + mass*a,3)} \\ \\mathrm{N}$.

b)

We resolve $F=ma$ in the upwards positive direction. The normal reaction force, $R$, will act in the opposite direction to the weight of the ball. This time however the box is decelerating so $a=\\var{-decelerate} \\, \\mathrm{ms^{-2}}$.

\\begin{align}
F & = ma \\\\
R - mg &= ma \\\\
R & = mg + ma \\\\
& = \\left(\\var{mass}\\times \\var{g}\\right) + \\left(\\var{mass} \\times \\var{-decelerate}\\right) \\\\
& = \\var{precround(mass*g - mass*decelerate,3)}.
\\end{align}

The normal reaction on the ball has magnitude $\\var{precround(mass*g - mass*decelerate,3)} \\, \\mathrm{N}$.

", "rulesets": {}, "parts": [{"precisionType": "dp", "prompt": "

What is the magnitude, in Newtons to 3 decimal places, of the normal reaction of the floor of the box on the ball?

", "precisionMessage": "You have not given your answer to the correct precision.", "allowFractions": false, "variableReplacements": [], "precision": "3", "maxValue": "mass*g + mass*a", "minValue": "mass*g + mass*a", "variableReplacementStrategy": "originalfirst", "strictPrecision": false, "correctAnswerFraction": false, "showCorrectAnswer": true, "precisionPartialCredit": 0, "scripts": {}, "marks": 1, "type": "numberentry", "showPrecisionHint": false}, {"precisionType": "dp", "prompt": "

The box then moves at a constant speed before decelerating to rest at $\\var{decelerate}\\ \\mathrm{ms^{-2}}$. What is the normal reaction, in Newtons to 3 decimal places, of the floor of the box to the ball during the time the box decelerates?

", "precisionMessage": "You have not given your answer to the correct precision.", "allowFractions": false, "variableReplacements": [], "precision": "3", "maxValue": "mass*g-mass*decelerate", "minValue": "mass*g-mass*decelerate", "variableReplacementStrategy": "originalfirst", "strictPrecision": false, "correctAnswerFraction": false, "showCorrectAnswer": true, "precisionPartialCredit": 0, "scripts": {}, "marks": 1, "type": "numberentry", "showPrecisionHint": false}], "statement": "

A ball of mass $\\var{mass}kg$ is in a box which accelerates upwards at a rate of $\\var{a} \\, \\mathrm{ms}^{-2}$.

", "variable_groups": [], "variablesTest": {"maxRuns": 100, "condition": ""}, "preamble": {"css": "", "js": ""}, "variables": {"a": {"definition": "random(0.25..3#0.25)", "templateType": "randrange", "group": "Ungrouped variables", "name": "a", "description": ""}, "decelerate": {"definition": "random(0.3..5#0.35)", "templateType": "randrange", "group": "Ungrouped variables", "name": "decelerate", "description": ""}, "mass": {"definition": "random(1..100#1)", "templateType": "randrange", "group": "Ungrouped variables", "name": "mass", "description": ""}, "g": {"definition": "9.8", "templateType": "number", "group": "Ungrouped variables", "name": "g", "description": ""}}, "metadata": {"description": "

A mass is inside a box which is suspended vertically by a cord. Question uses $F=ma$ for different accelerations.

a)

We resolve $F=ma$ with the positive direction heading upwards, as the ball is moving upwards. Both the resistance and the force due to gravity are acting in the downwards direction, against the path of the ball.

\\begin{align}
F & = ma \\\\
\\simplify[]{-{resistance} -({mass}*9.8)} &= \\var{mass} \\times a \\\\[0.5em]
a &= \\simplify[]{ (-{resistance}-({mass}*9.8))/{mass} } \\\\[0.5em]
&= \\var{precround(-acceleration,3)}
\\end{align}

We are asked to give the ball's deceleration, which is the opposite of its acceleration. The deceleration of the ball is $\\var{precround(acceleration,3)} \\ \\mathrm{ms^{-2}}$.

b)

When the ball reaches its greatest height it stops moving for an instant before it changes direction. At this instant, $v=0$. Now we have $u = \\var{u}, a = \\var{-precround(acceleration,3)}$ and $v=0$ and we want to find the distance travelled, $s$. We can use the formula $v^2 = u^2 + 2as$, rearranged for $s$.

\\begin{align}
v^2 & = u^2 + 2as \\\\
2as & = v^2 - u^2 \\\\
s & = \\frac{v^2 - u^2}{2a} \\\\[0.5em]
& = \\frac{ 0 - \\var{u}^2}{2 \\times (\\var{-precround(acceleration,3)})} \\\\[0.5em]
& = \\var{precround(max_height,3)}
\\end{align}

The greatest height reached by the ball is $\\var{precround(max_height,3)} \\ \\mathrm{m}$.

c)

To find the time taken to reach $\\var{s}\\, \\mathrm{m}$ we can use the equation $v = u + at$, where $v=0, u= \\var{u}$ and $a = \\var{-precround(acceleration,3)}$.

\\begin{align}
v & = u + at \\\\
0 & = \\var{u} - \\left(\\var{a} \\times t \\right) \\\\[0.5em]
t & = \\frac{\\var{u}}{\\var{a}} \\\\[0.5em]
& = \\var{precround(time_up,3)}
\\end{align}

The time taken for the ball to reach $\\var{s} \\, \\mathrm{m}$ from the bottom of the pool is $\\var{precround(time_up,3)}$ seconds.

d)

On the ball's descent back down to the bottom of the pool we can take vertically downwards as the positive direction, and resolve $F=ma$. The resistance of the water still acts against the ball's movement, but this time the force of gravity acts in the positive direction.

\\begin{align}
F & = ma \\\\
\\simplify[]{{mass}*9.8 - {resistance}} &= \\var{mass} \\times a \\\\[0.5em]
a &= \\simplify[]{({mass}*9.8-{resistance})/{mass}} \\\\[0.5em]
&= \\var{precround(acceleration2,3)}
\\end{align}

The acceleration of the ball on its descent back down to its initial position is $\\var{precround(acceleration2,3)} \\ \\mathrm{ms^{-2}}$.

e)

We have that $a = \\var{precround(acceleration2,3)}$ as the ball heads back down to the bottom of the pool. We can model this journey back down as a particle starting with initial velocity $u=0$: the ball 'starts' its journey downwards when it reaches its greatest height - where its speed is zero. We also have the distance it will travel downwards as $s=\\var{precround(max_height,3)}$. We want to know the final velocity $v$ when the ball hits the floor, so we can use the equation $v^2 = u^2 + 2as$.

\\begin{align}
v^2 & = u^2 + 2as \\\\
v^2 & = \\simplify[]{0^2+(2*{precround(acceleration2,3)}*{precround(max_height,3)})} \\\\
& = \\var{2*precround(acceleration2,3)*precround(max_height,3)}\\\\
v & = \\var{precround(v_bottom,3)}
\\end{align}

The speed of the ball when it hits the bottom of the pool on its return is $\\var{precround(v_bottom,3)} \\, \\mathrm{ms^{-1}}$.

f)

To find the time taken for the ball to return to the ground from the greatest height of $\\var{s} \\, \\mathrm{m}$ we can use the equation $s = ut + \\frac{1}{2}at^2$ with $s = \\var{precround(max_height,3)}, u = 0$ and $a = \\var{precround(acceleration2,3)}$.

\\begin{align}
s & = ut + \\frac{1}{2}at^2 \\\\
\\var{precround(max_height,3)} &= \\simplify[]{0 + (1/2)*({precround(acceleration2,3)}*t^2)}\\\\
t^2 & = \\frac{2 \\times \\var{precround(max_height,3)}}{\\var{precround(acceleration2,3)}} \\\\
& = \\var{2*max_height/acceleration2} \\\\
t & = \\var{precround(time_down,3)}
\\end{align}

The ball takes $\\var{precround(time_down,3)}$ seconds to return to the ground from its greatest height.

g)

The total time of the ball's journey (to reach the greatest height and fall back down) is found by adding the solutions to parts c and f.

\\begin{align} t & = \\var{precround(time_up,3)} + \\var{precround(time_down,3)} \\\\
& = \\var{precround(time_up+time_down,3)}
\\end{align}

The total time is $\\var{precround(time_up+time_down,3)}$ seconds.

", "rulesets": {}, "parts": [{"precisionType": "dp", "prompt": "

Find the vertical deceleration of the ball, in $\\mathrm{ms^{-2}}$ to 3 decimal places.

", "precisionMessage": "You have not given your answer to the correct precision.", "allowFractions": false, "variableReplacements": [], "maxValue": "acceleration", "strictPrecision": false, "minValue": "acceleration", "variableReplacementStrategy": "originalfirst", "precisionPartialCredit": 0, "correctAnswerFraction": false, "showCorrectAnswer": true, "precision": "3", "scripts": {}, "marks": 1, "type": "numberentry", "showPrecisionHint": false}, {"precisionType": "dp", "prompt": "

Find the greatest height that the ball reaches above the bottom of the pool, in $\\mathrm{m}$ to 3 decimal places.

", "precisionMessage": "

You have not given your answer to the correct precision.

", "allowFractions": false, "variableReplacements": [{"variable": "acceleration", "part": "p0", "must_go_first": false}], "maxValue": "max_height", "strictPrecision": false, "minValue": "max_height", "variableReplacementStrategy": "originalfirst", "precisionPartialCredit": 0, "correctAnswerFraction": false, "showCorrectAnswer": true, "precision": "3", "scripts": {}, "marks": 1, "type": "numberentry", "showPrecisionHint": false}, {"precisionType": "dp", "prompt": "

Find the time taken to reach this greatest height, in seconds to 3 decimal places.

", "precisionMessage": "

You have not given your answer to the correct precision.

", "allowFractions": false, "variableReplacements": [{"variable": "acceleration", "part": "p0", "must_go_first": false}], "maxValue": "time_up", "strictPrecision": false, "minValue": "time_up", "variableReplacementStrategy": "originalfirst", "precisionPartialCredit": 0, "correctAnswerFraction": false, "showCorrectAnswer": true, "precision": "3", "scripts": {}, "marks": 1, "type": "numberentry", "showPrecisionHint": false}, {"precisionType": "dp", "prompt": "

Find the acceleration in $\\mathrm{ms^{-2}}$ of the ball on its return down to the bottom of the pool (to 3d.p.).

", "precisionMessage": "You have not given your answer to the correct precision.", "allowFractions": false, "variableReplacements": [], "maxValue": "acceleration2", "strictPrecision": false, "minValue": "acceleration2", "variableReplacementStrategy": "originalfirst", "precisionPartialCredit": 0, "correctAnswerFraction": false, "showCorrectAnswer": true, "precision": "3", "scripts": {}, "marks": 1, "type": "numberentry", "showPrecisionHint": false}, {"precisionType": "dp", "prompt": "

Find the speed in $\\mathrm{ms^{-1}}$ of the ball as it hits the bottom of the pool floor on its return (to 3.d.p.).

", "precisionMessage": "

You have not given your answer to the correct precision.

", "allowFractions": false, "variableReplacements": [{"variable": "acceleration2", "part": "p3", "must_go_first": false}, {"variable": "max_height", "part": "p1", "must_go_first": false}], "maxValue": "sqrt(2*acceleration2*max_height)", "strictPrecision": false, "minValue": "sqrt(2*acceleration2*max_height)", "variableReplacementStrategy": "originalfirst", "precisionPartialCredit": 0, "correctAnswerFraction": false, "showCorrectAnswer": true, "precision": "3", "scripts": {}, "marks": 1, "type": "numberentry", "showPrecisionHint": false}, {"precisionType": "dp", "prompt": "

Find the time in seconds taken for the ball to drop from its greatest height back to the bottom of the pool (to 3d.p.).

", "precisionMessage": "

You have not given your answer to the correct precision.

", "allowFractions": false, "variableReplacements": [{"variable": "max_height", "part": "p1", "must_go_first": false}, {"variable": "acceleration2", "part": "p3", "must_go_first": false}], "maxValue": "time_down", "strictPrecision": false, "minValue": "time_down", "variableReplacementStrategy": "originalfirst", "precisionPartialCredit": 0, "correctAnswerFraction": false, "showCorrectAnswer": true, "precision": "3", "scripts": {}, "marks": 1, "type": "numberentry", "showPrecisionHint": false}, {"precisionType": "dp", "prompt": "

Find the total time in seconds taken for the ball to return to its initial position (to 3d.p.).

", "precisionMessage": "

You have not given your answer to the correct precision.

", "allowFractions": false, "variableReplacements": [{"variable": "time_up", "part": "p2", "must_go_first": false}, {"variable": "time_down", "part": "p5", "must_go_first": false}], "maxValue": "time_up+time_down", "strictPrecision": false, "minValue": "time_up+time_down", "variableReplacementStrategy": "originalfirst", "precisionPartialCredit": 0, "correctAnswerFraction": false, "showCorrectAnswer": true, "precision": "3", "scripts": {}, "marks": 1, "type": "numberentry", "showPrecisionHint": false}], "statement": "

With a speed of $\\var{u} \\ \\mathrm{ms^{-1}}$ a ball of mass $\\var{massg} \\ \\mathrm{g}$ is projected vertically upwards from the bottom of a full swimming pool. The ball experiences a constant resistance of $\\var{resistance} \\ \\mathrm{N}$ as it moves through the water. The ball never reaches the surface of the swimming pool.

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

", "variable_groups": [{"variables": ["g", "u", "resistance", "massg"], "name": "Setup"}], "variablesTest": {"maxRuns": 100, "condition": ""}, "variables": {"acceleration": {"definition": "(resistance+mass*g)/mass", "templateType": "anything", "group": "Ungrouped variables", "name": "acceleration", "description": "

Acceleration downwards, due to resistance and gravity.

"}, "time_up": {"definition": "u/acceleration", "templateType": "anything", "group": "Ungrouped variables", "name": "time_up", "description": "

Time taken to reach the maximum height.

"}, "g": {"definition": "9.8", "templateType": "number", "group": "Setup", "name": "g", "description": ""}, "resistance": {"definition": "random(2..5#0.5)", "templateType": "randrange", "group": "Setup", "name": "resistance", "description": "

resistance in water in Newtons

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

Maximum height reached by the ball.

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mass in grams

"}, "a2": {"definition": "precround(acceleration2,3)", "templateType": "anything", "group": "Ungrouped variables", "name": "a2", "description": "

acceleration of the ball as it returns to the pool floor, to 3dp

"}, "mass": {"definition": "massg/1000", "templateType": "anything", "group": "Ungrouped variables", "name": "mass", "description": "

mass in kg that will be used in solution.

"}, "time_down": {"definition": "sqrt(2*max_height/acceleration2)", "templateType": "anything", "group": "Ungrouped variables", "name": "time_down", "description": "

Time taken to sink back to the bottom

"}, "u": {"definition": "random(1..15#0.5)", "templateType": "randrange", "group": "Setup", "name": "u", "description": "

initial ball speed

"}, "acceleration2": {"definition": "(mass*g-resistance)/mass", "templateType": "anything", "group": "Ungrouped variables", "name": "acceleration2", "description": ""}}, "metadata": {"description": "

Ball projected vertically within water of a swimming pool. Find the acceleration, greatest height above the swimming pool floor reached, speed and time. Uses SUVAT equations and $F=ma$.

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