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This deck focuses on Design Momentum Conservation Experiments, giving you a quick way to review the definitions, rules, and examples that matter most for Physics.
Study Design Momentum Conservation Experiments in Physics with focused flashcards that help you recognize the idea, recall the key rule, and apply it in practice-style prompts.
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Find the total initial momentum if m1=0.50kg at +0.80m/s and m2=0.50kg at −0.20m/s.
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pi=0.30kg⋅m/s. pi=(0.50)(0.80)+(0.50)(−0.20)=0.40−0.10=0.30
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This deck focuses on Design Momentum Conservation Experiments, giving you a quick way to review the definitions, rules, and examples that matter most for Physics.
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.
Answer: pi=0.30kg⋅m/s. pi=(0.50)(0.80)+(0.50)(−0.20)=0.40−0.10=0.30
Answer: pi=m1v1i+m2v2i. Sum individual momenta algebraically with proper signs.
Answer: Choose one direction as positive; opposite direction is negative. Sign convention prevents errors when velocities oppose each other.
Answer: Perfectly inelastic collision (carts stick; share one final velocity). Single final velocity simplifies analysis.
Answer: Both carts (and any attached masses) treated as one system. System boundary must include all interacting objects for conservation to apply.
Answer: pf=0.16kg⋅m/s. pf=(0.40)(0.25)+(0.60)(0.10)=0.10+0.06=0.16
Answer: A cart at rest remains at rest (no drift) on the track. No drift indicates zero net force along track direction.
Answer: Level the track (minimizes external force component along motion). Eliminates gravity component along track, unlike fans or magnets which add forces.
Answer: Each cart mass m and velocity v before and after the interaction. Need all masses and velocities to calculate p=mv for each object.
Answer: Track level and friction (use low-friction track, level it). Minimizing friction and gravity components ensures negligible external forces.
Answer: Two carts on a straight track with labeled m1,m2 and v1i,v2i,v1f,v2f. Shows all needed quantities for momentum calculations.
Answer: pf=0.40kg⋅m/s. pf=(0.50)(0.20)+(0.30)(1.00)=0.10+0.30=0.40kg⋅m/s
Answer: Photogate(s) with a flag of known length (or motion sensor). Time through flag gives velocity; motion sensors measure directly.
Answer: Cart mass configuration (values of m1 and/or m2). You manipulate mass to see its effect on conservation.
Answer: Yes, within experimental uncertainty (pf≈pi). 3.3% difference is within typical experimental error.
Answer: One-dimensional head-on collision along a straight track. Simplifies vector analysis to scalar math with sign convention.
Answer: Level, low-friction track (or air track) with minimal contact friction. Minimizes external forces that would violate conservation.
Answer: Initial conditions (for example v1i or mass distribution). The experimenter controls what collides and how fast.
Answer: m1v1i+m2v2i=m1v1f+m2v2f. Total momentum before equals total momentum after in isolated systems.
Answer: p=mv. Momentum equals mass times velocity vector.
Answer: Negative momentum: p<0. Velocity opposite to positive direction has negative sign.
Answer: pi=0.40kg⋅m/s. pi=(0.50)(0.80)+(0.30)(0)=0.40kg⋅m/s
Answer: Change in system momentum: Δp=pf−pi (or percent difference). Measures how well momentum is conserved in the collision.
Answer: Repeat trials and average results (report spread/uncertainty). Multiple trials reduce random errors and reveal consistency.
Answer: Photogates or motion sensor (or video analysis with scale). These tools measure velocity precisely without disturbing the motion.
Answer: Verify constant velocity for a single cart: Δv≈0 over time. No velocity change confirms negligible external forces.
Answer: m1v1i+m2v2i=m1v1f+m2v2f. Total momentum before equals total momentum after for two objects.
Answer: pf=m1v1f+m2v2f. Sum final momenta to compare with initial total momentum.
Answer: %diff=∣pi∣∣pf−pi∣×100%. Normalizes difference by initial value for comparison.
Answer: Scatter plot of pf vs pi; ideal trend is pf=pi (slope 1). Perfect conservation shows as 45° line through origin.
Answer: Net external impulse is negligible: ∑FextΔt≈0. External forces must be minimal during collision time for momentum to be conserved.
Answer: Difference between totals: Δp=pf−pi (or percent difference). Measures how well momentum is conserved.
Answer: Track conditions (levelness and friction) and the collision mechanism. Controls ensure only mass variation affects results.
Answer: Flag length L to compute v=ΔtL. Photogate timing gives Δt; need L for velocity.
Answer: Use a short collision time and measure velocities immediately before and after. Minimizes impulse from friction during collision.
Answer: Net external impulse is zero, ∑Jext=0 (isolated system). No external forces means no external impulse to change total momentum.
Answer: p=mv. Momentum is mass times velocity, both magnitude and direction matter.
Answer: Choose + direction along the track; opposite motion has negative v. Ensures consistent velocity signs for momentum calculations.
Answer: 6%. %diff=0.50∣0.47−0.50∣×100%=6%
Answer: Percent difference: 2∣pf∣+∣pi∣∣pf−pi∣×100%. Percent difference accounts for measurement scale and uncertainty.