Physics Flashcards: Apply Physics To Collision Design

What this deck covers

This deck focuses on Apply Physics To Collision Design, giving you a quick way to review the definitions, rules, and examples that matter most for Physics.

How to use these flashcards

Work through these flashcards in short sessions. Try to answer each prompt before flipping the card, then revisit any cards you miss until the explanation feels automatic.

Flashcard 1: What quantity equals the loss of kinetic energy in an inelastic collision (as heat, sound, deformation)?

Answer: $\Delta K=K_{\text{after}}-K_{\text{before}}<0$. Energy is lost to heat, sound, and permanent deformation.

Flashcard 2: What is the impulse–momentum theorem relating impulse $\vec{J}$ to momentum change?

Answer: $\vec{J}=\Delta \vec{p}$. Impulse equals the change in momentum by Newton's second law.

Flashcard 3: Which design change reduces peak force for the same $\Delta p$: increase or decrease $\Delta t$?

Answer: Increase $\Delta t$. Since $\vec{J}=\vec{F}_{\text{avg}}\Delta t=\Delta \vec{p}$, larger $\Delta t$ reduces $\vec{F}_{\text{avg}}$.

Flashcard 4: In a 1D collision, what does $e=0$ indicate about the post-collision motion?

Answer: Objects move together: $v_{1f}=v_{2f}$. Zero restitution means perfectly inelastic collision.

Flashcard 5: Find $F_{\text{avg}}$ if $\Delta p=12\,\text{N}\cdot\text{s}$ and $\Delta t=0.06\,\text{s}$.

Answer: $F_{\text{avg}}=200\,\text{N}$. $F_{\text{avg}}=\frac{\Delta p}{\Delta t}=\frac{12}{0.06}=200\,\text{N}$.

Flashcard 6: Which option increases stopping distance and reduces peak force: rigid bumper or crumple zone?

Answer: Crumple zone. Crumple zones deform to extend collision time and reduce peak force.

Flashcard 7: State the formula for linear momentum of an object in collision analysis.

Answer: $\vec{p}=m\vec{v}$. Momentum equals mass times velocity vector.

Flashcard 8: Which collision type has $e=0$ with objects sticking together after impact?

Answer: Perfectly inelastic collision. Maximum energy loss, objects move together after impact.

Flashcard 9: Identify the correct relation between force and momentum change during a collision.

Answer: $\vec{F}_{\text{net}}=\frac{\Delta \vec{p}}{\Delta t}$. Average force equals momentum change divided by time interval.

Flashcard 10: What principle states that total momentum of an isolated collision system stays constant?

Answer: Conservation of momentum: $\sum \vec{p}_{\text{before}}=\sum \vec{p}_{\text{after}}$. In isolated systems, no external forces change the total momentum.

Flashcard 11: What design change reduces average impact force for the same momentum change?

Answer: Increase collision time $\Delta t$. Longer impact time reduces force by $F=\frac{\Delta p}{\Delta t}$.

Flashcard 12: Find the average deceleration magnitude if a car slows from $25$ to $0\,\text{m/s}$ in $0.50\,\text{s}$.

Answer: $|a|=50\,\text{m/s}^2$. $a=\frac{\Delta v}{\Delta t}=\frac{0-25}{0.50}=-50\,\text{m/s}^2$.

Flashcard 13: Find the stopping distance if a $1000\,\text{kg}$ car at $20\,\text{m/s}$ is stopped by $F_{\text{avg}}=2.0\times 10^5\,\text{N}$.

Answer: $d=1.0\,\text{m}$. $W=Fd=\Delta K$; $d=\frac{\frac{1}{2}mv^2}{F}=\frac{200000}{200000}=1.0\,\text{m}$.

Flashcard 14: State the kinetic energy formula used in collision energy calculations.

Answer: $K=\frac{1}{2}mv^2$. Kinetic energy is half mass times velocity squared.

Flashcard 15: A $2.0\,\text{kg}$ cart at $3.0\,\text{m/s}$ sticks to a $1.0\,\text{kg}$ cart at rest; find $v_f$.

Answer: $2.0\,\text{m/s}$. Using conservation: $(2.0)(3.0)+(1.0)(0)=(2.0+1.0)v_f$, so $v_f=2.0\,\text{m/s}$.

Flashcard 16: State the impulse-momentum theorem used to design collision cushioning.

Answer: $\vec{J}=\Delta \vec{p}=\vec{F}_{\text{avg}}\Delta t$. Impulse equals momentum change and force times time.

Flashcard 17: 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.

Answer: $\Delta K=6\,\text{J}$ lost. $K_i=8\,\text{J}$, $K_f=2\,\text{J}$, so $\Delta K=-6\,\text{J}$.

Flashcard 18: If the stopping distance is doubled at the same initial speed, how does average force change?

Answer: Average force halves: $F_{\text{avg}}\propto \frac{1}{d}$. Work-energy theorem: $F_{\text{avg}}d=\Delta K$, so doubling $d$ halves $F_{\text{avg}}$.

Flashcard 19: Which collision type has $e=1$ and conserves kinetic energy (idealized)?

Answer: Perfectly elastic collision. No energy lost, objects bounce apart with $e=1$.

Flashcard 20: Which option best reduces peak force in a bumper: larger $\Delta t$ or smaller $\Delta t$ for the same $\Delta p$?

Answer: Larger $\Delta t$. Larger $\Delta t$ reduces force since $F=\frac{\Delta p}{\Delta t}$.

Flashcard 21: What is the work–energy relation used to estimate average stopping force over distance $d$?

Answer: $F_{\text{avg}}d=\Delta K$. Work done by stopping force equals change in kinetic energy.

Flashcard 22: What collision type has objects stick together and share a final velocity?

Answer: Perfectly inelastic collision. Maximum kinetic energy is lost when objects stick together.

Flashcard 23: 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.

Answer: $v_f=2\,\text{m/s}$. $(1+2)v_f=1(6)+2(0)$; $3v_f=6$; $v_f=2\,\text{m/s}$.

Flashcard 24: State the formula for linear momentum of an object moving with velocity $\vec{v}$.

Answer: $\vec{p}=m\vec{v}$. Momentum equals mass times velocity, a vector quantity.

Flashcard 25: State the coefficient of restitution formula using relative speeds along the line of impact.

Answer: $e=\frac{|v_{2f}-v_{1f}|}{|v_{1i}-v_{2i}|}$. Ratio of relative separation speed to relative approach speed.

Flashcard 26: Find $e$ if approach speed is $10\,\text{m/s}$ and separation speed is $3\,\text{m/s}$ along the impact line.

Answer: $e=0.30$. $e=\frac{\text{separation speed}}{\text{approach speed}}=\frac{3}{10}=0.30$.

Flashcard 27: What is the kinetic energy formula used to check energy changes in collisions?

Answer: $K=\frac{1}{2}mv^2$. Kinetic energy is half mass times velocity squared.

Flashcard 28: State the work-energy relation often used to size a crumple zone distance.

Answer: $W=\Delta K$. Work done equals the change in kinetic energy.

Flashcard 29: What is the coefficient of restitution $e$ in terms of relative speeds along the impact line?

Answer: $e=\frac{|v_{2f}-v_{1f}|}{|v_{1i}-v_{2i}|}$. Ratio of separation to approach speeds along impact line.

Flashcard 30: Find the impulse when a $2\,\text{kg}$ cart changes velocity from $+3$ to $-1\,\text{m/s}$ in 1D.

Answer: $\vec{J}=\Delta \vec{p}=-8\,\text{N}\cdot\text{s}$. $\Delta p=m\Delta v=2((-1)-(+3))=-8\,\text{N}\cdot\text{s}$.

Flashcard 31: In a force–time graph, what physical quantity equals the area under the curve during impact?

Answer: Impulse $\vec{J}$. Area under force-time curve equals impulse by definition.

Flashcard 32: What collision type conserves both momentum and kinetic energy (ideal case)?

Answer: Perfectly elastic collision. No kinetic energy is lost to heat or deformation in ideal elastic collisions.

Flashcard 33: Two equal masses collide head-on elastically; one is initially at rest. What happens to their speeds?

Answer: They exchange speeds (moving one stops). For equal masses in elastic collision, velocities are exchanged.

Flashcard 34: State the formula for impulse from an average force $\vec{F}_{\text{avg}}$ over time $\Delta t$.

Answer: $\vec{J}=\vec{F}_{\text{avg}}\Delta t$. Impulse is the product of average force and time duration.

Flashcard 35: State the 1D momentum conservation equation for two objects with masses $m_1,m_2$ and speeds $v$.

Answer: $m_1v_{1i}+m_2v_{2i}=m_1v_{1f}+m_2v_{2f}$. Total momentum before equals total momentum after in 1D collisions.

Flashcard 36: What condition must be met to treat a collision system as isolated for momentum conservation?

Answer: Net external impulse is negligible: $\vec{J}_{\text{ext}}\approx 0$. External forces like friction must be negligible compared to collision forces.

Flashcard 37: What condition must be true to treat a collision system as isolated for momentum analysis?

Answer: Net external impulse is negligible: $\vec{J}_{\text{ext}}\approx 0$. External forces must be negligible compared to collision forces.

Flashcard 38: 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.

Answer: $v_f=1\,\text{m/s}$. $(m_1+m_2)v_f=m_1v_1$; $4v_f=4$; $v_f=1\,\text{m/s}$.