4 - Momentum, Conservation, Impulse | Lecture Note

University:
High School
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High School Physics
Academic year:

2024
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9
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4: Momentum, Conservation, Impulse AP FIZZ Motivation • We can conserve energy for single objects with and without friction. Are there other quantities that can be conserved to analyze the motion of systems of objects as the collide with an without friction? Momentum • Vector quantity; product of mass and velocity, rate of change of net force 𝑝Ԧ = 𝑚𝑣Ԧ • Unit: kg·m/s Conservation of Momentum (1D & 2D) •For system of particles: • (total momentum)_initial = (total momentum)_final (all directions) •𝑝𝑖 = 𝑝𝑓 EX 1: 1.0kg cart travelling at 6.0m/s collides and sticks to a stationary 2.0kg cart. What speed do the connected carts have after the collision? 𝑝𝑖 = 𝑚1 𝑣1𝑖 + 𝑚2 𝑣2𝑖 𝑝𝑖 = 1.0 6.0 + 2.0 0 𝑝𝑖 = 6.0𝑘𝑔 · 𝑚/𝑠 𝑝𝑓 = 𝑚1 𝑣1𝑓 + 𝑚2 𝑣2𝑓 𝑝𝑓 = (𝑚1 +𝑚2 )𝑣𝑓 6.0 = (1.0 + 2.0)𝑣𝑓 𝑣𝑓 = 2.0 𝑚/𝑠 EX 2: A white ball (0.500kg) is shot at (4.00m/s) in a straight line at a stationary black ball (0.400kg). If the white ball deflects away from the line at 30.0° at a speed of 2.80m/s, what is the speed and angle of the black ball? 𝑝𝑖 = 0.500 4.00 + (0.400)(0) 𝑚 𝑝𝑖𝑥 = 2.00𝑘𝑔 · , 𝑝𝑖𝑦 = 0𝑘𝑔 · 𝑚/𝑠 𝑠 𝑝𝑓𝑥 = 𝑚1𝑣1𝑓𝑥 + 𝑚2𝑣2𝑓𝑥 2.00 = 0.500 2.80 cos 30.0 + 0.400 𝑣2𝑓 cos θ 2.00 = 1.2124 + 0.400 𝑣2𝑓 cos θ 1.9689 = 𝑣2𝑓 cos θ EX 2: A white ball (0.500kg) is shot at (4.00m/s) in a straight line at a stationary black ball (0.400kg). If the white ball deflects away from the line at 30.0° at a speed of 2.80m/s, what is the speed and angle of the black ball? 𝑝𝑓𝑦 = 𝑚1𝑣1𝑓𝑦 + 𝑚2𝑣2𝑓𝑦 0.00 = 0.500 2.80 sin 30.0 − 0.400 𝑣2𝑓 sin θ 0.00 = 0.700 − 0.400 𝑣2𝑓 sin θ 1.75 = 𝑣2𝑓 sin θ 1.9689 = 𝑣2𝑓 cos θ 𝜽 = 𝟒𝟏. 𝟔°, 𝒗𝟐𝒇 = 𝟐. 𝟔𝟑𝒎/𝒔 Impulse (J) • Change in momentum = impulse (same unit as momentum) • Rate of change of momentum = impulse/time = net force 𝐽 = Δ𝑝 = 𝑚𝑣2 − 𝑚𝑣1 𝐽 Δ𝑝 = = 𝐹𝑛𝑒𝑡 Δ𝑡 Δ𝑡 EX 3: What is the average horizontal force acting on a ball when it elastically bounces off a wall, assuming the collision time is 0.1 s and the momentum of the ball before collision was 2.0kg·m/s, perpendicular to the wall? If momentum before = 2.0kg·m/s, then momentum after = -2.0kg·m/s (opposite direction) Δ𝑝 𝐹𝑛𝑒𝑡 = Δ𝑡 −2.0 − (2.0) 𝐹𝑛𝑒𝑡 = 0.1 𝐹𝑛𝑒𝑡 = −40𝑁 (40N in direction after rebound)