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Collisions

Collisions. Collisions. Abelardo M. Zerda III Michael O. Suarez Jm Dawn C. Rivas Leslie Kate Diane Berte. Subtopics. Momentum and Impulse Conservation of Momentum Collisions in One Dimensions Collisions in Two Dimensions. Momentum.

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Collisions

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  1. Collisions Collisions Abelardo M. Zerda III Michael O. Suarez Jm Dawn C. Rivas Leslie Kate Diane Berte

  2. Subtopics • Momentum and Impulse • Conservation of Momentum • Collisions in One Dimensions • Collisions in Two Dimensions

  3. Momentum • Measure of one’s motion, equivalent to the product of one’s mass and velocity. • Momentum involves motion and mass. • Expressing the definition mathematically, p = mv where: p = momentum in kg-m/s units m = mass in kg units v = velocity of the object in m/s units

  4. Sample Problem • Calculate the momentum of a 100-kg missile traveling at 100 m/s eastward. • A 100 kg car strikes a tree at 50 km/h. Find its momentum.

  5. Impulse • Impulse – force (F) applied by one object to another object within a given time interval Δt • The formula to get impulse is: F Δt = Δp where: F Δt = impulse in N-s (newton-second) units Δp = change in momentum in kg-m/s units

  6. Sample Problem • A person in a sled with a total mass of 125 kg, slides down a grassy hill and reach a speed of 8 m/s at the bottom. If a pile of grass can exert a constant force of 250 newtons, how fast will the sled stop? • A 100 kg car strikes a tree at 20 km/h and comes to a stop in 0.5 seconds. Find its initial momentum and the force on the car while it is being stopped.

  7. Conservation of Momentum • In a collision, energy is not always conserved but all collisions have to conserve momentum if there is no net applied force. ∑ p before collision = ∑ p after collision

  8. Collisions in One Dimension There are three types of collisions: • Elastic collisions – conserve kinetic energy • Inelastic collisions – do not conserve kinetic energy • Completely inelastic collision

  9. Elastic Collisions in One Dimension • Let us assume that for the two balls, one is running on a straight track toward the second one, which is stationary. Assume the following items are given: m1 – mass of object 1 m2 – mass of object 2 v1 – velocity of object 1 before collision v2 – velocity of object 2 before collision v1’ – velocity of object 1 after collision v2’ – velocity of object 2 after collision

  10. Solving for the velocity of mass 1 after collision v1’: v1’ = [(m1 – m2) / (m1 + m2)] v1 • Solving for the velocity of mass 2 after collision v2’: v2’ = [2m1 / (m1 + m2)] v1

  11. Case 1: Both masses are equal v1’ = [(m1 – m2) / (m1 + m2)] v1 = [(m-m) / (m+m)] v1 = 0 After colliding with m2, m1 stops. v2’ = [2m1 / (m1 + m2)] v1 = (2m / 2m) v1 = v1 Total energy transfer occurred between m1 and m2.

  12. Case 2: Mass of moving object greater than the mass of the stationary object • Since m1 > m2, the term (m1 – m2) is positive. v1’ = [(m1 – m2) / (m1 + m2)] v1 m1 still moves and v1’ has the same direction as v1. • Since all the terms are nonzero then, v2’ = [2m1 / (m1 + m2)] v1 m2 also moves and v2’ has the same direction as v1

  13. Case 3: Mass of moving object less than the mass of the stationary object • Since m1 < m2, the term (m1 – m2) is negative. v1’ = [(m1 – m2) / (m1 + m2)] v1 m1 still moves but in the opposite direction. v2’ = [2m1 / (m1 + m2)] v1 Since all the terms are nonzero, then m2 also moves and v2 has the same direction as v1.

  14. Total Inelastic Collisions in One Dimension • Inelastic collisions produce a combined mass after collision. • The equation for this type of collision is: m1 v1 + m2 v2 = (m1 + m2) vf where: m1 + m2 is the combined mass of the object Vf is the final velocity value for the combined mass If m2 v2 = 0, vf = v1 m1/ (m1 + m2)

  15. Collisions in Two Dimensions • Balanced momentum in the x-direction and y-direction • Angles are arbitrary

  16. Using the Conservation of Energy: • Along the x-axis: px before collision = px after collision p1cos 1 + p2cos 2 = p1cos 1 + p2cos 2 • Along the y-axis: py before collision = pyafter collision p1sin 1 + p2sin 2 = p1sin 1 + p2sin 2

  17. Sample Problem: • A pool ball weighing 2 kg is traveling at 30 at 0.8 m/s hits another ball moving at 0.5 m/s at 180. If the second ball leaves the collision at 0 and the first moves away at 150, find the final velocity vectors of the balls.

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