Optimize Designs for Collision Safety - Physics
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Which option gives the smaller average force for the same $\Delta p$: $\Delta t=0.10,\text{s}$ or $\Delta t=0.40,\text{s}$?
Which option gives the smaller average force for the same $\Delta p$: $\Delta t=0.10,\text{s}$ or $\Delta t=0.40,\text{s}$?
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$\Delta t=0.40,\text{s}$. Longer time interval reduces average force per impulse-momentum theorem.
$\Delta t=0.40,\text{s}$. Longer time interval reduces average force per impulse-momentum theorem.
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What is the impulse-momentum theorem relating impulse to change in momentum?
What is the impulse-momentum theorem relating impulse to change in momentum?
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$J=\Delta p=F_{\text{avg}}\Delta t$. Impulse equals momentum change and also equals average force times time interval.
$J=\Delta p=F_{\text{avg}}\Delta t$. Impulse equals momentum change and also equals average force times time interval.
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What equation links average impact force to change in momentum and collision time?
What equation links average impact force to change in momentum and collision time?
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$F_{\text{avg}}=\frac{\Delta p}{\Delta t}$. Rearranging impulse-momentum theorem to solve for average force.
$F_{\text{avg}}=\frac{\Delta p}{\Delta t}$. Rearranging impulse-momentum theorem to solve for average force.
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What is the momentum of an object of mass $m$ moving at speed $v$?
What is the momentum of an object of mass $m$ moving at speed $v$?
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$p=mv$. Momentum is the product of mass and velocity.
$p=mv$. Momentum is the product of mass and velocity.
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What is the pressure if a force of $2000,\text{N}$ acts over an area of $0.020,\text{m}^2$?
What is the pressure if a force of $2000,\text{N}$ acts over an area of $0.020,\text{m}^2$?
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$1.0\times10^5,\text{Pa}$. Using $P=\frac{F}{A}=\frac{2000}{0.020}$.
$1.0\times10^5,\text{Pa}$. Using $P=\frac{F}{A}=\frac{2000}{0.020}$.
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What is the translational kinetic energy of a mass $m$ moving at speed $v$?
What is the translational kinetic energy of a mass $m$ moving at speed $v$?
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$K=\frac{1}{2}mv^2$. Kinetic energy equals half the mass times velocity squared.
$K=\frac{1}{2}mv^2$. Kinetic energy equals half the mass times velocity squared.
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Which design change most directly reduces peak force for the same momentum change: increase or decrease $\Delta t$?
Which design change most directly reduces peak force for the same momentum change: increase or decrease $\Delta t$?
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Increase $\Delta t$. Longer collision time reduces average force for same momentum change.
Increase $\Delta t$. Longer collision time reduces average force for same momentum change.
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Which collision generally produces a larger peak force: elastic or inelastic, for similar $\Delta t$?
Which collision generally produces a larger peak force: elastic or inelastic, for similar $\Delta t$?
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Elastic. Elastic collisions have higher rebound velocities, creating larger momentum changes.
Elastic. Elastic collisions have higher rebound velocities, creating larger momentum changes.
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What is the coefficient of restitution in terms of relative speeds along the line of impact?
What is the coefficient of restitution in terms of relative speeds along the line of impact?
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$e=\frac{v_{\text{sep}}}{v_{\text{app}}}$. Ratio of separation to approach speeds measures collision elasticity.
$e=\frac{v_{\text{sep}}}{v_{\text{app}}}$. Ratio of separation to approach speeds measures collision elasticity.
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What is the work-energy relation that connects stopping distance $d$ to average stopping force?
What is the work-energy relation that connects stopping distance $d$ to average stopping force?
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$W=F_{\text{avg}}d=\Delta K$. Work done by average force equals change in kinetic energy.
$W=F_{\text{avg}}d=\Delta K$. Work done by average force equals change in kinetic energy.
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What formula gives average deceleration when an object stops from speed $v$ over distance $d$?
What formula gives average deceleration when an object stops from speed $v$ over distance $d$?
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$a_{\text{avg}}=\frac{v^2}{2d}$. Derived from kinematic equation $v^2 = v_0^2 + 2ad$ with final velocity zero.
$a_{\text{avg}}=\frac{v^2}{2d}$. Derived from kinematic equation $v^2 = v_0^2 + 2ad$ with final velocity zero.
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What formula gives average stopping force for mass $m$ stopping from speed $v$ over distance $d$?
What formula gives average stopping force for mass $m$ stopping from speed $v$ over distance $d$?
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$F_{\text{avg}}=\frac{mv^2}{2d}$. Combines $F=ma$ with $a=\frac{v^2}{2d}$ from kinematics.
$F_{\text{avg}}=\frac{mv^2}{2d}$. Combines $F=ma$ with $a=\frac{v^2}{2d}$ from kinematics.
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Identify the safest choice: for fixed $m$ and $v$, should a design maximize or minimize crush distance $d$?
Identify the safest choice: for fixed $m$ and $v$, should a design maximize or minimize crush distance $d$?
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Maximize $d$. Larger crush distance reduces average force, improving safety.
Maximize $d$. Larger crush distance reduces average force, improving safety.
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Which option reduces pressure on contact for the same force: increase or decrease contact area $A$?
Which option reduces pressure on contact for the same force: increase or decrease contact area $A$?
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Increase $A$. Larger contact area distributes force, reducing pressure.
Increase $A$. Larger contact area distributes force, reducing pressure.
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What is the pressure formula relating force to contact area?
What is the pressure formula relating force to contact area?
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$P=\frac{F}{A}$. Pressure equals force divided by contact area.
$P=\frac{F}{A}$. Pressure equals force divided by contact area.
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What is $F_{\text{avg}}$ if $\Delta p=6000,\text{kg}\cdot\text{m/s}$ and $\Delta t=0.20,\text{s}$?
What is $F_{\text{avg}}$ if $\Delta p=6000,\text{kg}\cdot\text{m/s}$ and $\Delta t=0.20,\text{s}$?
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$3.0\times10^4,\text{N}$. Direct application of $F_{avg}=\frac{\Delta p}{\Delta t}$.
$3.0\times10^4,\text{N}$. Direct application of $F_{avg}=\frac{\Delta p}{\Delta t}$.
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What is $F_{\text{avg}}$ if a $1200,\text{kg}$ car stops from $15,\text{m/s}$ in $0.30,\text{s}$?
What is $F_{\text{avg}}$ if a $1200,\text{kg}$ car stops from $15,\text{m/s}$ in $0.30,\text{s}$?
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$6.0\times10^4,\text{N}$. Using $F_{avg}=\frac{m\Delta v}{\Delta t}=\frac{1200(15)}{0.30}$.
$6.0\times10^4,\text{N}$. Using $F_{avg}=\frac{m\Delta v}{\Delta t}=\frac{1200(15)}{0.30}$.
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What is $F_{\text{avg}}$ if a $1500,\text{kg}$ car stops from $20,\text{m/s}$ over $2.0,\text{m}$?
What is $F_{\text{avg}}$ if a $1500,\text{kg}$ car stops from $20,\text{m/s}$ over $2.0,\text{m}$?
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$1.5\times10^5,\text{N}$. Using $F_{avg}=\frac{mv^2}{2d}=\frac{1500(20)^2}{2(2.0)}$.
$1.5\times10^5,\text{N}$. Using $F_{avg}=\frac{mv^2}{2d}=\frac{1500(20)^2}{2(2.0)}$.
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What is $F_{\text{avg}}$ if a $1500,\text{kg}$ car stops from $20,\text{m/s}$ over $4.0,\text{m}$?
What is $F_{\text{avg}}$ if a $1500,\text{kg}$ car stops from $20,\text{m/s}$ over $4.0,\text{m}$?
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$7.5\times10^4,\text{N}$. Using $F_{avg}=\frac{mv^2}{2d}=\frac{1500(20)^2}{2(4.0)}$.
$7.5\times10^4,\text{N}$. Using $F_{avg}=\frac{mv^2}{2d}=\frac{1500(20)^2}{2(4.0)}$.
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State the impulse formula in terms of average force $F_{avg}$ and collision time $\Delta t$.
State the impulse formula in terms of average force $F_{avg}$ and collision time $\Delta t$.
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$J=F_{avg}\Delta t$. Impulse is the product of average force and time duration.
$J=F_{avg}\Delta t$. Impulse is the product of average force and time duration.
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What is the impulse–momentum theorem relating impulse $J$ to momentum change $\Delta p$?
What is the impulse–momentum theorem relating impulse $J$ to momentum change $\Delta p$?
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$J=\Delta p$. Impulse equals the change in momentum for any collision.
$J=\Delta p$. Impulse equals the change in momentum for any collision.
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Which collision type has the smallest rebound and typically reduces peak forces: more elastic or more inelastic?
Which collision type has the smallest rebound and typically reduces peak forces: more elastic or more inelastic?
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More inelastic. Inelastic collisions absorb more energy, reducing rebound.
More inelastic. Inelastic collisions absorb more energy, reducing rebound.
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For the same impulse $J$, which force profile has the lower peak force: longer, lower force or shorter, higher force?
For the same impulse $J$, which force profile has the lower peak force: longer, lower force or shorter, higher force?
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Longer, lower force. Extended time spreads impulse, reducing peak force.
Longer, lower force. Extended time spreads impulse, reducing peak force.
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In a collision, which quantity is the area under a force–time graph: impulse $J$ or kinetic energy $K$?
In a collision, which quantity is the area under a force–time graph: impulse $J$ or kinetic energy $K$?
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Impulse $J$. Area under F-t graph gives total impulse delivered.
Impulse $J$. Area under F-t graph gives total impulse delivered.
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State the average force in a collision in terms of momentum change $\Delta p$ and time $\Delta t$.
State the average force in a collision in terms of momentum change $\Delta p$ and time $\Delta t$.
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$F_{avg}=\frac{\Delta p}{\Delta t}$. Rearranged impulse-momentum theorem solving for force.
$F_{avg}=\frac{\Delta p}{\Delta t}$. Rearranged impulse-momentum theorem solving for force.
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What is the kinetic energy formula used to estimate crash energy that must be absorbed?
What is the kinetic energy formula used to estimate crash energy that must be absorbed?
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$K=\frac{1}{2}mv^2$. Kinetic energy depends on mass and velocity squared.
$K=\frac{1}{2}mv^2$. Kinetic energy depends on mass and velocity squared.
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State the work–energy relation for energy absorbed by a stopping force over distance $d$.
State the work–energy relation for energy absorbed by a stopping force over distance $d$.
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$W=\Delta K$. Work done equals the change in kinetic energy.
$W=\Delta K$. Work done equals the change in kinetic energy.
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For fixed crash energy $\Delta K$, which reduces average stopping force: larger $d$ or smaller $d$?
For fixed crash energy $\Delta K$, which reduces average stopping force: larger $d$ or smaller $d$?
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Larger $d$. Greater stopping distance reduces force for same energy.
Larger $d$. Greater stopping distance reduces force for same energy.
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State the pressure equation relating force $F$ and contact area $A$.
State the pressure equation relating force $F$ and contact area $A$.
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$P=\frac{F}{A}$. Pressure is force divided by contact area.
$P=\frac{F}{A}$. Pressure is force divided by contact area.
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State the constant-force work formula used for crash stopping distance $d$.
State the constant-force work formula used for crash stopping distance $d$.
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$W=Fd$. Work equals force times distance for constant force.
$W=Fd$. Work equals force times distance for constant force.
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