// Numbas version: exam_results_page_options {"name": "Dynamics - Impulse and momentum", "duration": 0, "metadata": {"description": "

To be used on the Mechanics wiki page under the Dynamics section, impulse and momentum page. Questions about impulse, momentum and collisions.

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Momentum $p \\ \\mathrm{Ns} = $ mass $\\mathrm{kg} \\ \\times $ velocity $\\mathrm{ms^{-1}}$.

\n

Remember that in the above formula, the mass is measured in $\\mathrm{kg}$.

\n

a)

\n

We have

\n

\\begin{align}
p & = \\frac{\\var{mass1}}{1000} \\times \\var{velocity1}, \\\\
&= \\var{precround(kg1*velocity1,3)} \\, \\mathrm{Ns}.
\\end{align}

\n

The magnitude of the momentum is $\\var{precround(kg1*velocity1,3)} \\, \\mathrm{Ns}.$

\n

b)

\n

We have

\n

\\begin{align}
p & = (\\var{tonnes1} \\times 1000) \\times \\var{velocity2}, \\\\
& = \\var{precround(tonnes1*1000*velocity2,3)} \\, \\mathrm{Ns}.
\\end{align}

\n

The magnitude of the momentum is $\\var{precround(tonnes1*1000*velocity2,3)} \\, \\mathrm{Ns}.$

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

A particle of mass $\\var{mass1} \\mathrm{g}$ moving at $\\var{velocity1} \\mathrm{ms^{-1}}$.

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A caravan of mass $\\var{tonnes1} \\ \\mathrm{tonnes}$ moving at $\\var{velocity2} \\mathrm{ms^{-1}}$.

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Find the magnitude of the momentum, $p \\ \\mathrm{Ns}$, of the following objects. Give your answers to 3 decimal places.

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mass in grams

"}, "velocity2": {"definition": "random(0.2..1.6#0.05)", "templateType": "randrange", "group": "Ungrouped variables", "name": "velocity2", "description": ""}, "velocity1": {"definition": "random(2..20#1)", "templateType": "randrange", "group": "Ungrouped variables", "name": "velocity1", "description": ""}, "tonnes1": {"definition": "random(0.5..10#0.5)", "templateType": "randrange", "group": "Ungrouped variables", "name": "tonnes1", "description": ""}}, "metadata": {"description": "

Find the magnitude of the momentum of some objects, given mass and speed. Apply the formula $\\rho = mv$.

", "licence": "Creative Commons Attribution 4.0 International"}, "type": "question"}, {"name": "Find change in speed after force applied for given time", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "contributors": [{"name": "Amy Chadwick", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/505/"}], "functions": {}, "ungrouped_variables": ["mass", "force", "time", "impulse"], "tags": [], "advice": "

a)

\n

The impulse imparted is given by $\\text{impulse} = \\text{force} \\times \\text{time}$.

\n

So we have $\\var{force} \\times \\var{time} = \\var{impulse} \\mathrm{Ns}.$

\n

b)

\n

The Impulse-Momentum Principle states that $\\text{Impulse} = \\text{Final momentum} - \\text{Initial momentum}$.

\n

So we have

\n

\\begin{align}
\\var{impulse} & = \\var{mass}v - \\var{mass}u, \\\\
& = \\var{mass}v - 0, \\\\
v & = \\frac{\\var{impulse}}{\\var{mass}}, \\\\
& =\\var{impulse/mass} \\mathrm{ms^{-1}}.
\\end{align}

\n

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

To 3 significant figures, find the magnitude of the impulse given to the body by the force (in $\\mathrm{Ns}$).

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

To 3 significant figures, find the final speed of the body in $\\mathrm{ms^{-1}}$.

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A body of mass $\\var{mass} \\mathrm{kg}$ is initially at rest on a smooth horizontal plane. Suppose that a horizontal force of magnitude $\\var{force} \\mathrm{N}$ acts on the body for $\\var{time} \\mathrm{s}$. 

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Use the Impulse-Momentum principle to find the change in speed of an object after a constant force is applied for a given period of time.

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a)

\n

Impulse is change in momentum. The ball begins at rest so initially has a momentum of zero.

\n

Therefore

\n

\\begin{align}
I & = \\var{football_mass} \\times\\var{football_speed} - 0, \\\\
& = \\var{siground(football_mass*football_speed,3)} \\, \\mathrm{Ns}.
\\end{align}

\n

The impulse received by the ball is $ \\var{siground(football_mass*football_speed,3)} \\, \\mathrm{Ns}$.

\n

b)

\n

Impulse is change in momentum. Therefore we take the rebound direction as being positive and have that 

\n

\\begin{align}
I & = mv - mu, \\\\
\\var{squash_impulse}  & = \\var{squash_mass} v - (\\var{squash_mass} \\times \\var{- squash_speed}), \\\\
v & = \\frac{\\var{squash_impulse} + (\\var{squash_mass} \\times \\var{-squash_speed})}{\\var{squash_mass}}, \\\\
& = \\var{siground((squash_impulse + squash_mass*-squash_speed)/squash_mass,3)} \\, \\mathrm{ms^{-1}}.
\\end{align}

\n

The speed after the ball rebounds is $\\var{siground(squash_second_speed,3)} \\, \\mathrm{ms^{-1}}$.

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A football of mass $\\var{football_mass} \\, \\mathrm{kg}$ is at rest before it is kicked.

\n

What is the impulse received by the ball in $\\mathrm{Ns}$ if its speed is $\\var{football_speed} \\, \\mathrm{ms^{-1}}$ after it is struck?

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A squash ball of mass $\\var{squash_mass} \\, \\mathrm{kg}$ hits a fixed vertical wall at right angles with speed $\\var{squash_speed} \\, \\mathrm{ms^{-1}}$. The ball rebounds at right angles to the wall.

\n

What is the speed of the ball in $\\mathrm{ms^{-1}}$ just after it has hit the wall, given that the magnitude of the impulse exerted by the wall on the ball is $\\var{squash_impulse} \\, \\mathrm{Ns}$?

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Give your answers to the following questions to 3 significant figures.

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coefficient of restitution of the wall

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

Finding speeds and impulses 

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We can draw a diagram with all the speeds and impulses with arrows.

\n

\n

Here particle $A$ has mass $m_1 = \\var{A_mass} \\, \\mathrm{kg}$ and is projected with a speed $u_1 \\, \\mathrm{ms^{-1}} = \\var{A_speed} \\, \\mathrm{ms^{-1}}$. Particle $B$ has mass $m_2 \\, \\mathrm{kg}$ to be determined and is initially at rest: $u_2 = 0 \\, \\mathrm{ms^{-1}}$. Both particles have common final velocities $v = \\var{particle_speed} \\, \\mathrm{ms^{-1}}$ and exert an impulse $I \\, \\mathrm{Ns}$ through the string as shown.

\n

a)

\n

To find the mass of particle $B$ we can use the principle of conservation of momentum: the total momentum in a system remains constant.

\n

\\begin{align}
m_1 u_1 + m_2 u_2 & = m_1 v_1 + m_2v_2, \\\\
(\\var{A_mass} \\times \\var{A_speed}) + 0 & = \\var{particle_speed} ( \\var{A_mass} + m_2 ), \\\\
\\frac{ \\var{A_mass} \\times \\var{A_speed}}{\\var{particle_speed}} & =  \\var{A_mass} + m_2, \\\\
m_2  & = \\frac{ \\var{A_mass} \\times \\var{A_speed}}{\\var{particle_speed}} - \\var{A_mass}, \\\\
& = \\var{siground( (A_mass*A_speed)/particle_speed - A_mass,3)} \\, \\mathrm{kg}.
\\end{align}

\n

So the mass of particle $B$ is $\\var{siground( (A_mass*A_speed)/particle_speed - A_mass,3)} \\, \\mathrm{kg}$.

\n

b)

\n

To find the impulse we consider one of the particles and apply the Impulse-Momentum Principle. Although it is easier to consider particle $B$ as it is initially at rest we will consider particle $A$ incase the mass of particle $B$ was calculated incorrectly in part a).

\n

We will consider particle $A$ and apply the Impulse-Momentum Principle in the direction of the impulse shown in the diagram, $(\\leftarrow)$. This means our velocities will now be negative as they are acting in the opposite direction to the impulse.

\n

\\begin{align}
I & = mv - mu_1, \\\\
& = m (v - u_1), \\\\
& = \\var{A_mass} \\times ( - \\var{particle_speed} - ( - \\var{A_speed})), \\\\
& = \\var{A_mass} \\times ( \\var{ - particle_speed + A_speed} ), \\\\
& = \\var{siground( A_mass*(A_speed - particle_speed),3)} \\, \\mathrm{Ns}.
\\end{align}

\n

The magnitude of the impulse transmitted through the string is $\\var{impulse} \\, \\mathrm{Ns}$.

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

If the mass of particle $A$ is $\\var{A_mass} \\, \\mathrm{kg}$, what is the mass of particle $B$ in $\\mathrm{kg}$?

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Find the magnitude in $\\mathrm{Ns}$ of the impulse transmitted through the string when it goes taut.

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Two particles $A$ and $B$ are at rest and connected by a light inextensible string which is slack. Particle $A$ is projected directly away from particle $B$ with speed $\\var{A_speed} \\mathrm{ms^{-1}}$.

\n

When the string goes taut the common speed of the particles is $\\var{particle_speed} \\mathrm{ms^{-1}}$.

\n

Give your answers to the following questions to 3 significant figures.

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common speed of particles when the string goes taut

"}, "A_speed": {"definition": "random(5.1..7#0.1)", "templateType": "randrange", "group": "Ungrouped variables", "name": "A_speed", "description": "

particle A is projected away from B with this speed

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mass $A$

"}, "impulse": {"definition": "-A_mass*(particle_speed-A_speed)", "templateType": "anything", "group": "Ungrouped variables", "name": "impulse", "description": ""}}, "metadata": {"description": "

Collisions question involving principle of conservation of momentum.

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We can draw a diagram.

\n

\n

The impulse applied to both balls is $I = \\var{impulse} \\, \\mathrm{Ns}$, and the speeds after the collision are $v_1 = \\var{A_speed} \\, \\mathrm{ms^{-1}}$ and $v_2 = \\var{B_speed} \\, \\mathrm{ms^{-1}}$ for the balls $A$ and $B$ respectively.

\n

a)

\n

To calculate $u_1$, the speed of ball $A$ before the collision, we use the equation $I = m_1v_1 - m_1u_1$, where $m_1$ is the mass of ball $A$, which is $\\var{A_mass} \\, \\mathrm{kg}$. We resolve in the direction of the impulse, therefore the signs of $v_1$ and $u_1$ will be reversed.

\n

\\begin{align}
I & = m_1v_1 - m_1u_1, \\\\
\\var{impulse} & = \\var{A_mass} (v_1 - u_1), \\\\
\\frac{\\var{impulse}}{\\var{A_mass}} & = - \\var{A_speed} - (- u_1), \\\\
u_1 & = \\frac{\\var{impulse}}{\\var{A_mass}} + \\var{A_speed}, \\\\
& = \\var{siground( impulse/A_mass + A_speed,3)} \\, \\mathrm{ms^{-1}}.
\\end{align}

\n

The magnitude of the speed of $A$ before the collision is $\\var{siground( impulse/A_mass + A_speed,3)} \\, \\mathrm{ms^{-1}}$.

\n

b) 

\n

To calculate $u_2$, the speed of ball $B$ before the collision we use the equation $I = m_2v_2 - m_2u_2$, where $m_2$ is the mass of ball $B$, which is $\\var{B_mass} \\, \\mathrm{kg}$. We resolve in the direction of the impulse shown in the diagram, which is the same as the direction of the speeds.

\n

\\begin{align}
I & = m_2v_2 - m_2u_2, \\\\
\\var{impulse} & = \\var{B_mass} ( \\var{B_speed} - u_2), \\\\
\\frac{\\var{impulse} }{ \\var{B_mass}} & = \\var{B_speed} - u_2, \\\\
u_2 & = \\var{B_speed} -  \\frac{\\var{impulse} }{ \\var{B_mass}}, \\\\
& = \\var{siground( B_speed - (impulse/B_mass),3)} \\, \\mathrm{ms^{-1}}.
\\end{align}

\n

If this is positive it means the direction of ball $B$ we assumed is correct; if it is negative it means the ball was originally travelling in the other direction. However we were asked to find the magnitude of the speed so we take our answer as $\\var{B_before_speed} \\, \\mathrm{ms^{-1}}$.

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

Find the magnitude of the speed of $A$ before the collision, in $\\mathrm{ms^{-1}}$ to 3 s.f.

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Find the magnitude of the speed of $B$ before the collision, in $\\mathrm{ms^{-1}}$ to 3 s.f.

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

Two balls $A$ and $B$ of mass $\\var{A_mass} \\mathrm{kg}$ and $\\var{B_mass} \\mathrm{kg}$ respectively are moving in the same straight line on a smooth horizontal surface. The balls collide. After the collision both of the balls are moving in the same direction. 

\n

After the collision, the $A$ travels at $\\var{A_speed} \\mathrm{ms^{-1}}$ and $B$ travels at $\\var{B_speed} \\mathrm{ms^{-1}}$. The magnitude of the impulse of $A$ on $B$ is $\\var{impulse} \\mathrm{Ns}$.

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Two particles collide. Given the masses and final speeds of the particles, as well as the impulse imparted on each, find their initial speeds.

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