Apply Physics to Collision Design - Physics
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Find the kinetic energy lost when $m_1=1,\text{kg}$ at $4,\text{m/s}$ sticks to $m_2=3,\text{kg}$ at rest.
Find the kinetic energy lost when $m_1=1,\text{kg}$ at $4,\text{m/s}$ sticks to $m_2=3,\text{kg}$ at rest.
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$\Delta K=6,\text{J}$ lost. $K_i=8,\text{J}$, $K_f=2,\text{J}$, so $\Delta K=-6,\text{J}$.
$\Delta K=6,\text{J}$ lost. $K_i=8,\text{J}$, $K_f=2,\text{J}$, so $\Delta K=-6,\text{J}$.
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Find the final velocity of a $2,\text{kg}$ cart initially at rest hit by a $1,\text{kg}$ cart at $6,\text{m/s}$; they stick.
Find the final velocity of a $2,\text{kg}$ cart initially at rest hit by a $1,\text{kg}$ cart at $6,\text{m/s}$; they stick.
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$v_f=2,\text{m/s}$. $(1+2)v_f=1(6)+2(0)$; $3v_f=6$; $v_f=2,\text{m/s}$.
$v_f=2,\text{m/s}$. $(1+2)v_f=1(6)+2(0)$; $3v_f=6$; $v_f=2,\text{m/s}$.
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Find the common final velocity when $m_1=1,\text{kg}$ at $4,\text{m/s}$ sticks to $m_2=3,\text{kg}$ at rest.
Find the common final velocity when $m_1=1,\text{kg}$ at $4,\text{m/s}$ sticks to $m_2=3,\text{kg}$ at rest.
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$v_f=1,\text{m/s}$. $(m_1+m_2)v_f=m_1v_1$; $4v_f=4$; $v_f=1,\text{m/s}$.
$v_f=1,\text{m/s}$. $(m_1+m_2)v_f=m_1v_1$; $4v_f=4$; $v_f=1,\text{m/s}$.
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Find $F_{\text{avg}}$ if $\Delta p=12,\text{N}\cdot\text{s}$ and $\Delta t=0.06,\text{s}$.
Find $F_{\text{avg}}$ if $\Delta p=12,\text{N}\cdot\text{s}$ and $\Delta t=0.06,\text{s}$.
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$F_{\text{avg}}=200,\text{N}$. $F_{\text{avg}}=\frac{\Delta p}{\Delta t}=\frac{12}{0.06}=200,\text{N}$.
$F_{\text{avg}}=200,\text{N}$. $F_{\text{avg}}=\frac{\Delta p}{\Delta t}=\frac{12}{0.06}=200,\text{N}$.
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Find the impulse when a $2,\text{kg}$ cart changes velocity from $+3$ to $-1,\text{m/s}$ in 1D.
Find the impulse when a $2,\text{kg}$ cart changes velocity from $+3$ to $-1,\text{m/s}$ in 1D.
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$\vec{J}=\Delta \vec{p}=-8,\text{N}\cdot\text{s}$. $\Delta p=m\Delta v=2((-1)-(+3))=-8,\text{N}\cdot\text{s}$.
$\vec{J}=\Delta \vec{p}=-8,\text{N}\cdot\text{s}$. $\Delta p=m\Delta v=2((-1)-(+3))=-8,\text{N}\cdot\text{s}$.
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Identify the correct relation between force and momentum change during a collision.
Identify the correct relation between force and momentum change during a collision.
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$\vec{F}_{\text{net}}=\frac{\Delta \vec{p}}{\Delta t}$. Average force equals momentum change divided by time interval.
$\vec{F}_{\text{net}}=\frac{\Delta \vec{p}}{\Delta t}$. Average force equals momentum change divided by time interval.
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Which collision type has $e=0$ with objects sticking together after impact?
Which collision type has $e=0$ with objects sticking together after impact?
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Perfectly inelastic collision. Maximum energy loss, objects move together after impact.
Perfectly inelastic collision. Maximum energy loss, objects move together after impact.
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What condition must be true to treat a collision system as isolated for momentum analysis?
What condition must be true to treat a collision system as isolated for momentum analysis?
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Net external impulse is negligible: $\vec{J}_{\text{ext}}\approx 0$. External forces must be negligible compared to collision forces.
Net external impulse is negligible: $\vec{J}_{\text{ext}}\approx 0$. External forces must be negligible compared to collision forces.
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Which collision type has $e=1$ and conserves kinetic energy (idealized)?
Which collision type has $e=1$ and conserves kinetic energy (idealized)?
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Perfectly elastic collision. No energy lost, objects bounce apart with $e=1$.
Perfectly elastic collision. No energy lost, objects bounce apart with $e=1$.
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What principle states that total momentum of an isolated collision system stays constant?
What principle states that total momentum of an isolated collision system stays constant?
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Conservation of momentum: $\sum \vec{p}{\text{before}}=\sum \vec{p}{\text{after}}$. In isolated systems, no external forces change the total momentum.
Conservation of momentum: $\sum \vec{p}{\text{before}}=\sum \vec{p}{\text{after}}$. In isolated systems, no external forces change the total momentum.
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What is the coefficient of restitution $e$ in terms of relative speeds along the impact line?
What is the coefficient of restitution $e$ in terms of relative speeds along the impact line?
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$e=\frac{|v_{2f}-v_{1f}|}{|v_{1i}-v_{2i}|}$. Ratio of separation to approach speeds along impact line.
$e=\frac{|v_{2f}-v_{1f}|}{|v_{1i}-v_{2i}|}$. Ratio of separation to approach speeds along impact line.
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State the formula for linear momentum of an object in collision analysis.
State the formula for linear momentum of an object in collision analysis.
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$\vec{p}=m\vec{v}$. Momentum equals mass times velocity vector.
$\vec{p}=m\vec{v}$. Momentum equals mass times velocity vector.
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State the impulse-momentum theorem used to design collision cushioning.
State the impulse-momentum theorem used to design collision cushioning.
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$\vec{J}=\Delta \vec{p}=\vec{F}_{\text{avg}}\Delta t$. Impulse equals momentum change and force times time.
$\vec{J}=\Delta \vec{p}=\vec{F}_{\text{avg}}\Delta t$. Impulse equals momentum change and force times time.
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What design change reduces average impact force for the same momentum change?
What design change reduces average impact force for the same momentum change?
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Increase collision time $\Delta t$. Longer impact time reduces force by $F=\frac{\Delta p}{\Delta t}$.
Increase collision time $\Delta t$. Longer impact time reduces force by $F=\frac{\Delta p}{\Delta t}$.
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State the work-energy relation often used to size a crumple zone distance.
State the work-energy relation often used to size a crumple zone distance.
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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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State the kinetic energy formula used in collision energy calculations.
State the kinetic energy formula used in collision energy calculations.
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$K=\frac{1}{2}mv^2$. Kinetic energy is half mass times velocity squared.
$K=\frac{1}{2}mv^2$. Kinetic energy is half mass times velocity squared.
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Which option best reduces peak force in a bumper: larger $\Delta t$ or smaller $\Delta t$ for the same $\Delta p$?
Which option best reduces peak force in a bumper: larger $\Delta t$ or smaller $\Delta t$ for the same $\Delta p$?
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Larger $\Delta t$. Larger $\Delta t$ reduces force since $F=\frac{\Delta p}{\Delta t}$.
Larger $\Delta t$. Larger $\Delta t$ reduces force since $F=\frac{\Delta p}{\Delta t}$.
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Find $e$ if approach speed is $10,\text{m/s}$ and separation speed is $3,\text{m/s}$ along the impact line.
Find $e$ if approach speed is $10,\text{m/s}$ and separation speed is $3,\text{m/s}$ along the impact line.
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$e=0.30$. $e=\frac{\text{separation speed}}{\text{approach speed}}=\frac{3}{10}=0.30$.
$e=0.30$. $e=\frac{\text{separation speed}}{\text{approach speed}}=\frac{3}{10}=0.30$.
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Find the average deceleration magnitude if a car slows from $25$ to $0,\text{m/s}$ in $0.50,\text{s}$.
Find the average deceleration magnitude if a car slows from $25$ to $0,\text{m/s}$ in $0.50,\text{s}$.
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$|a|=50,\text{m/s}^2$. $a=\frac{\Delta v}{\Delta t}=\frac{0-25}{0.50}=-50,\text{m/s}^2$.
$|a|=50,\text{m/s}^2$. $a=\frac{\Delta v}{\Delta t}=\frac{0-25}{0.50}=-50,\text{m/s}^2$.
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Find the stopping distance if a $1000,\text{kg}$ car at $20,\text{m/s}$ is stopped by $F_{\text{avg}}=2.0\times 10^5,\text{N}$.
Find the stopping distance if a $1000,\text{kg}$ car at $20,\text{m/s}$ is stopped by $F_{\text{avg}}=2.0\times 10^5,\text{N}$.
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$d=1.0,\text{m}$. $W=Fd=\Delta K$; $d=\frac{\frac{1}{2}mv^2}{F}=\frac{200000}{200000}=1.0,\text{m}$.
$d=1.0,\text{m}$. $W=Fd=\Delta K$; $d=\frac{\frac{1}{2}mv^2}{F}=\frac{200000}{200000}=1.0,\text{m}$.
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Two equal masses collide head-on elastically; one is initially at rest. What happens to their speeds?
Two equal masses collide head-on elastically; one is initially at rest. What happens to their speeds?
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They exchange speeds (moving one stops). For equal masses in elastic collision, velocities are exchanged.
They exchange speeds (moving one stops). For equal masses in elastic collision, velocities are exchanged.
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What condition must be met to treat a collision system as isolated for momentum conservation?
What condition must be met to treat a collision system as isolated for momentum conservation?
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Net external impulse is negligible: $\vec{J}_{\text{ext}}\approx 0$. External forces like friction must be negligible compared to collision forces.
Net external impulse is negligible: $\vec{J}_{\text{ext}}\approx 0$. External forces like friction must be negligible compared to collision forces.
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In a force–time graph, what physical quantity equals the area under the curve during impact?
In a force–time graph, what physical quantity equals the area under the curve during impact?
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Impulse $\vec{J}$. Area under force-time curve equals impulse by definition.
Impulse $\vec{J}$. Area under force-time curve equals impulse by definition.
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Which option increases stopping distance and reduces peak force: rigid bumper or crumple zone?
Which option increases stopping distance and reduces peak force: rigid bumper or crumple zone?
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Crumple zone. Crumple zones deform to extend collision time and reduce peak force.
Crumple zone. Crumple zones deform to extend collision time and reduce peak force.
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If the stopping distance is doubled at the same initial speed, how does average force change?
If the stopping distance is doubled at the same initial speed, how does average force change?
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Average force halves: $F_{\text{avg}}\propto \frac{1}{d}$. Work-energy theorem: $F_{\text{avg}}d=\Delta K$, so doubling $d$ halves $F_{\text{avg}}$.
Average force halves: $F_{\text{avg}}\propto \frac{1}{d}$. Work-energy theorem: $F_{\text{avg}}d=\Delta K$, so doubling $d$ halves $F_{\text{avg}}$.
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What is the work–energy relation used to estimate average stopping force over distance $d$?
What is the work–energy relation used to estimate average stopping force over distance $d$?
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$F_{\text{avg}}d=\Delta K$. Work done by stopping force equals change in kinetic energy.
$F_{\text{avg}}d=\Delta K$. Work done by stopping force equals change in kinetic energy.
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State the 1D momentum conservation equation for two objects with masses $m_1,m_2$ and speeds $v$.
State the 1D momentum conservation equation for two objects with masses $m_1,m_2$ and speeds $v$.
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$m_1v_{1i}+m_2v_{2i}=m_1v_{1f}+m_2v_{2f}$. Total momentum before equals total momentum after in 1D collisions.
$m_1v_{1i}+m_2v_{2i}=m_1v_{1f}+m_2v_{2f}$. Total momentum before equals total momentum after in 1D collisions.
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What is the kinetic energy formula used to check energy changes in collisions?
What is the kinetic energy formula used to check energy changes in collisions?
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$K=\frac{1}{2}mv^2$. Kinetic energy is half mass times velocity squared.
$K=\frac{1}{2}mv^2$. Kinetic energy is half mass times velocity squared.
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A $2.0,\text{kg}$ cart at $3.0,\text{m/s}$ sticks to a $1.0,\text{kg}$ cart at rest; find $v_f$.
A $2.0,\text{kg}$ cart at $3.0,\text{m/s}$ sticks to a $1.0,\text{kg}$ cart at rest; find $v_f$.
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$2.0,\text{m/s}$. Using conservation: $(2.0)(3.0)+(1.0)(0)=(2.0+1.0)v_f$, so $v_f=2.0,\text{m/s}$.
$2.0,\text{m/s}$. Using conservation: $(2.0)(3.0)+(1.0)(0)=(2.0+1.0)v_f$, so $v_f=2.0,\text{m/s}$.
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What collision type conserves both momentum and kinetic energy (ideal case)?
What collision type conserves both momentum and kinetic energy (ideal case)?
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Perfectly elastic collision. No kinetic energy is lost to heat or deformation in ideal elastic collisions.
Perfectly elastic collision. No kinetic energy is lost to heat or deformation in ideal elastic collisions.
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