Physics Flashcards: Apply Momentum Conservation To Collisions

Study Apply Momentum Conservation To Collisions in Physics with focused flashcards that help you recognize the idea, recall the key rule, and apply it in practice-style prompts.

Physics

Apply Momentum Conservation To Collisions

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QUESTION
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What is the impulse-momentum theorem for an object in 1D?

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ANSWER

J=Δp=m(vu)J = \Delta p = m(v-u). Impulse equals change in momentum, which is mass times velocity change.

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Flashcard 1: What is the impulse-momentum theorem for an object in 1D?

Answer: J=Δp=m(vu)J = \Delta p = m(v-u). Impulse equals change in momentum, which is mass times velocity change.

Flashcard 2: What is the momentum conservation equation for a 1D collision in an isolated system?

Answer: m1u1+m2u2=m1v1+m2v2m_1u_1 + m_2u_2 = m_1v_1 + m_2v_2. Total momentum before equals total momentum after in isolated systems.

Flashcard 3: Identify the sign convention rule for 1D momentum problems along a chosen axis.

Answer: Choose + direction; velocities opposite it are negative. Consistent signs ensure momentum conservation equations work correctly.

Flashcard 4: What component equations express 2D momentum conservation for two objects?

Answer: px,i=px,fp_{x,i}=p_{x,f} and py,i=py,fp_{y,i}=p_{y,f}. Each component is independently conserved.

Flashcard 5: What is the 2D momentum conservation statement for an isolated collision?

Answer: pinitial=pfinal\vec p_{\text{initial}}=\vec p_{\text{final}}. Vector momentum is conserved in all directions.

Flashcard 6: What is the post-collision common velocity vv if two masses stick together in 1D?

Answer: v=m1u1+m2u2m1+m2v = \frac{m_1u_1 + m_2u_2}{m_1 + m_2}. Total momentum divided by total mass gives common velocity.

Flashcard 7: Find v2v_2 if m1=m2m_1 = m_2, u2=0u_2 = 0, and after collision v1=0v_1 = 0 (use momentum conservation).

Answer: v2=u1v_2 = u_1. Equal masses exchange velocities in this special case.

Flashcard 8: What value of ee corresponds to a perfectly inelastic collision (objects stick)?

Answer: e=0e = 0. Objects stick together with no relative motion after collision.

Flashcard 9: Find vv if m1=1kgm_1=1\,\text{kg} at u1=8m/su_1=8\,\text{m/s} sticks to m2=3kgm_2=3\,\text{kg} at u2=0u_2=0.

Answer: v=2m/sv=2\,\text{m/s}. (1)(8)+(3)(0)=(1+3)v(1)(8)+(3)(0)=(1+3)v gives v=2v=2 m/s.

Flashcard 10: Find vv if m1=3kgm_1=3\,\text{kg} at u1=2m/su_1=2\,\text{m/s} sticks to m2=3kgm_2=3\,\text{kg} at u2=2m/su_2=-2\,\text{m/s}.

Answer: v=0m/sv=0\,\text{m/s}. Equal and opposite momenta cancel when objects stick.

Flashcard 11: What value of ee corresponds to a perfectly inelastic collision (objects stick together)?

Answer: e=0e=0. Objects stick together with no relative motion after collision.

Flashcard 12: What condition must be true for total momentum to be conserved during a collision?

Answer: Fext,net=0F_{\text{ext,net}} = 0 (impulse from external forces is zero). No external forces means the system's total momentum stays constant.

Flashcard 13: What is the momentum-conservation equation for an isolated 1D collision of two objects?

Answer: m1u1+m2u2=m1v1+m2v2m_1u_1+m_2u_2=m_1v_1+m_2v_2. Total momentum before equals total momentum after for isolated systems.

Flashcard 14: Find the final velocity vv if m1=1kgm_1 = 1\,\text{kg} at u1=4m/su_1 = 4\,\text{m/s} hits m2=3kgm_2 = 3\,\text{kg} at u2=0u_2 = 0 and they stick.

Answer: v=1msv = 1\,\frac{\text{m}}{\text{s}}. (1)(4)+(3)(0)=(1+3)v(1)(4) + (3)(0) = (1+3)v gives v=1v = 1 m/s.

Flashcard 15: Identify the sign rule for 1D momentum problems when choosing a positive direction.

Answer: Velocities opposite the positive direction are negative. Ensures correct vector addition in 1D problems.

Flashcard 16: What is the coefficient of restitution definition in 1D for two colliding objects?

Answer: e=v2v1u1u2e=\frac{v_2-v_1}{u_1-u_2}. Measures relative separation speed over approach speed.

Flashcard 17: What is the momentum of a particle of mass mm moving at velocity vv in 1D?

Answer: p=mvp=mv. Momentum equals mass times velocity in classical mechanics.

Flashcard 18: What value of ee corresponds to a perfectly elastic collision?

Answer: e=1e=1. Objects separate with same relative speed as approach.

Flashcard 19: Find the final velocity vv if m1=2kgm_1 = 2\,\text{kg} at u1=3m/su_1 = 3\,\text{m/s} hits m2=1kgm_2 = 1\,\text{kg} at rest and they stick.

Answer: v=2m/sv = 2\,\text{m/s}. (2)(3)+(1)(0)=(2+1)v(2)(3) + (1)(0) = (2+1)v gives v=2v = 2 m/s.

Flashcard 20: Find vv if m1=2kgm_1=2\,\text{kg} at u1=3m/su_1=3\,\text{m/s} sticks to m2=1kgm_2=1\,\text{kg} at u2=0u_2=0.

Answer: v=2m/sv=2\,\text{m/s}. (2)(3)+(1)(0)=(2+1)v(2)(3)+(1)(0)=(2+1)v gives v=2v=2 m/s.

Flashcard 21: Find the impulse on a 2kg2\,\text{kg} cart that changes velocity from 1m/s-1\,\text{m/s} to 4m/s4\,\text{m/s}.

Answer: J=10NsJ=10\,\text{N}\cdot\text{s}. J=mΔv=(2)(4(1))=10J=m\Delta v=(2)(4-(-1))=10 N·s.

Flashcard 22: Find the recoil speed vrv_r of a 2kg2\,\text{kg} gun if it fires a 0.010kg0.010\,\text{kg} bullet at 400m/s400\,\text{m/s} from rest.

Answer: vr=2m/sv_r = -2\,\text{m/s}. (2)(0)+(0.010)(0)=(2)vr+(0.010)(400)(2)(0) + (0.010)(0) = (2)v_r + (0.010)(400) gives vr=2v_r = -2 m/s.

Flashcard 23: Identify the correct 2D conservation equations if total initial momentum is only in +x+x direction.

Answer: px,i=px,fp_{x,i}=p_{x,f} and 0=py,f0=p_{y,f}. py,i=0p_{y,i}=0 so py,f=0p_{y,f}=0; x-momentum conserved separately.

Flashcard 24: What is the impulse-momentum theorem for one object in 1D?

Answer: J=rianglep=m(vu)J= riangle p=m(v-u). Impulse equals change in momentum.

Flashcard 25: What additional conservation law applies to an elastic collision besides momentum?

Answer: Kinetic energy is conserved: Ki=KfK_i = K_f. Elastic collisions conserve both momentum and kinetic energy.

Flashcard 26: Which quantity is always conserved in an isolated collision: momentum or kinetic energy?

Answer: Momentum. KE only conserved in elastic collisions; momentum always conserved.

Flashcard 27: Find the system total momentum pp for m1=2kgm_1 = 2\,\text{kg} at +3m/s+3\,\text{m/s} and m2=1kgm_2 = 1\,\text{kg} at 2m/s-2\,\text{m/s}.

Answer: p=4kgm/sp = 4\,\text{kg}\cdot\text{m/s}. (2)(3)+(1)(2)=62=4(2)(3) + (1)(-2) = 6 - 2 = 4 kg·m/s.

Flashcard 28: What is the kinetic energy of a mass mm moving at speed vv?

Answer: K=12mv2K = \frac{1}{2}mv^2. Energy of motion equals half mass times velocity squared.

Flashcard 29: State the formula for the common final velocity when two objects stick together in 1D.

Answer: v=m1u1+m2u2m1+m2v=\frac{m_1u_1+m_2u_2}{m_1+m_2}. Total momentum divided by total mass gives common velocity.

Flashcard 30: What condition must be true for total momentum to be conserved during a collision?

Answer: Net external impulse on the system is 00. No external forces means momentum stays constant.

Flashcard 31: What is the momentum of a particle of mass mm moving at velocity vv in 1D?

Answer: p=mvp = mv. Momentum equals mass times velocity in the chosen direction.

Flashcard 32: What value of ee corresponds to a perfectly elastic collision?

Answer: e=1e = 1. Objects separate at same relative speed they approached.

Flashcard 33: Find the impulse JJ on a 0.50kg0.50\,\text{kg} ball if it changes from +4m/s+4\,\text{m/s} to 2m/s-2\,\text{m/s}.

Answer: J=3NsJ = -3\,\text{N}\cdot\text{s}. J=(0.50)[(2)(+4)]=3J = (0.50)[(-2) - (+4)] = -3 N·s.

Flashcard 34: Find the final speed vv if a 0.20kg0.20\,\text{kg} cart at 5m/s5\,\text{m/s} sticks to a 0.30kg0.30\,\text{kg} cart at rest.

Answer: v=2m/sv = 2\,\text{m/s}. (0.20)(5)+(0.30)(0)=(0.50)v(0.20)(5) + (0.30)(0) = (0.50)v gives v=2v = 2 m/s.