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Aim: How can we solve collision problems? Do Now: An object of mass m traveling at velocity 3v collides with an object of mass 2m traveling with velocity.

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Presentation on theme: "Aim: How can we solve collision problems? Do Now: An object of mass m traveling at velocity 3v collides with an object of mass 2m traveling with velocity."— Presentation transcript:

1 Aim: How can we solve collision problems? Do Now: An object of mass m traveling at velocity 3v collides with an object of mass 2m traveling with velocity –v. What is the combined initial momentum of the system? p i = p 1 + p 2 p i = mv + mv p i = m(3v) + 2m(-v) p i = 3mv – 2mv p i = mv HW #12 due tomorrow

2 Newton’s cradle demo How can we explain what is going on? The momentum of: one goes in, one goes out two goes in, two goes out etc.

3 The Law of Conservation of Momentum The momentum of any closed, isolated system does not change.

4

5 Elastic Collision Collision where the objects bounce off each other and kinetic energy is conserved (ie K i = K f )

6 Inelastic Collision A collision where kinetic energy is not conserved after the collision (ie K i > K f )

7 1. A tennis ball of mass m rebounds from a racquet with the same speed v as it had initially, as shown. The magnitude of the momentum change of the ball is (A)0 (B)mv (C)2mv (D)2mvsin  (E)2mvcos  Δp = m Δv Δp = m(v – v o ) Δp = mv – mv 0 Δp = -mvcosθ – (mvcosθ) Δp = -mvcosθ + (-mvcosθ) Δp = -2mvcosθ (magnitude = ignore the signs) No Calculator **1 minute 15 seconds**

8 2. Two bodies of masses 5 and 7 kilograms are initially at rest on a horizontal frictionless surface. A light spring is compressed between the bodies, which are held together by a thin thread. After the spring is released by burning through the thread, the 5 kilogram body has a speed of 0.2 m/s. The speed of the 7 kilogram body is (in m/s) (A) (B) (C) (D) (E) 5 kg7 kg mv = mv (5 kg)(0.2 m/s) = (7 kg)v 1 = 7 v v = 1/7 No Calculator **1 minute 15 seconds**

9 3. Two pucks are attached by a stretched spring and are initially held at rest on a frictionless surface, as shown above. The pucks are then released simultaneously. If puck I has three times the mass of puck II, which of the following quantities is the same for both pucks as the spring pulls the two pucks toward each other? (A)Speed (B)Velocity (C)Acceleration (D)Kinetic energy (E)Magnitude of momentum No Calculator **1 minute** Momentum is conserved

10 4. Two objects having the same mass travel toward each other on a flat surface, each with a speed of 1.0 meter per second relative to the surface. The objects collide head ‑ on and are reported to rebound after the collision, each with a speed of 2.0 meters per second relative to the surface. Which of the following assessments of this report is most accurate? (A)Momentum was not conserved, therefore the report is false. (B)If potential energy was released to the objects during the collision, the report could be true. (C)If the objects had different masses, the report could be true. (D)If the surface was inclined, the report could be true. (E)If there was no friction between the objects and the surface, the report could be true. No Calculator **1 minute** Release potential energy, such as a spring or explosion, could increase the speed after a collision

11 5. Two objects of mass 0.2 kg and 0.1 kg, respectively, move parallel to the x ‑ axis, as shown above. The 0.2 kg object overtakes and collides with the 0.1 kg object. Immediately after the collision, the y ‑ component of the velocity of the 0.2 kg object is 1 m/s upward. What is the y ‑ component of the velocity of the 0.1 kg object immediately after the collision? (A)2 m/s downward (B)0.5 m/s downward (C)0 m/s (D)0.5 m/s upward (E)2 m/s upward Δp = 0 p f – p i = 0 p i = p f 0 = mv + mv 0 = (0.2 kg)(1 m/s) + (0.1 kg)v -0.1v = 0.2 v = -2 m/s or 2 m/s downward No Calculator **1 minute 15 seconds**

12 6.Two people of unequal mass are initially standing still on ice with negligible friction. They then simultaneously push each other horizontally. Afterward, which of the following is true? (A)The kinetic energies of the two people are equal. (B)The speeds of the two people are equal. (C)The momentum of the two people are of equal magnitude. (D)The center of mass of the two ‑ person system moves in the direction of the less massive person. (E)The less massive person has a smaller initial acceleration than the more massive person. No Calculator **1 minute** Law of conservation of momentum

13 7. A stationary object explodes, breaking into three pieces of masses m, m, and 3m. The two pieces of mass m move off at right angles to each other with the same magnitude of momentum mV, as shown in the diagram above. What are the magnitude and direction of the velocity of the piece having mass 3m ? MagnitudeDirection (A) (B) (C) (D) (E) 3mv’ No Calculator **1 minute 15 sec**

14 8. An unstable nucleus, initially at rest, decays into two spherical fragments. Fragment A has radius r A charge q A, and rest mass m A ; fragment B has radius r B, charge q B, and rest mass m B. Suppose that after the decay process fragment A has velocity v A that is small ‑ compared with the velocity of light. Develop an expression for the kinetic energy of fragment B in terms of v A, m A, and m B. Calculator **2 minutes**

15 9. A massless spring is between a 1 kg mass and a 3 kg mass as shown above, but is not attached to either mass. Both masses are on a horizontal frictionless table. In an experiment, the 1 kg mass is held in place and the spring is compressed by pushing on the 3 kg mass. The 3 kg mass is then released and moves off with a speed of 10 meters per second. a. Determine the minimum work needed to compress the spring in this experiment. W = ΔK W = ½ mv 2 W = ½ (3 kg)(10 m/s) 2 W = 150 J Calculator **6 minutes**

16 The spring is compressed again exactly as before, but this time both masses are released simultaneously. b. Determine the final velocity of each mass relative to the cable after the masses are released. p 1 = p 2 m 1 v 1 = m 2 v 2 v 1 = m 2 v 2 /m 1 v 1 = (3 kg)v 2 /1 kg v 1 = 3 v 2 v 1 = 3(5 m/s) v 1 = 15 m/s K 1 + K 2 = E T = W ½ m 1 v ½ m 2 v 2 2 = W ½ m 1 (3v 2 ) 2 + ½ m 2 v 2 2 = W ½ (1 kg)(3v 2 ) 2 + ½ (3 kg)v 2 2 = 150 J 4.5v v 2 2 = 150 J 6v 2 2 = 150 J v 2 2 = 25 v 2 = 5 m/s

17 10. An incident ball A of mass 0.10 kg is sliding at 1.4 m/s on the horizontal tabletop of negligible friction shown above. It makes a head ‑ on collision with a target ball B of mass 0.50 kg at rest at the edge of the table. As a result of the collision, the incident ball rebounds, sliding backwards at 0.70 m/s immediately after the collision. Calculator **15 minutes**

18 a. Calculate the speed of the 0.50 kg target ball immediately after the collision.

19 The tabletop is 1.20 m above a level, horizontal floor. The target ball is projected horizontally and initially strikes the floor at a horizontal displacement d from the point of collision. b. Calculate the horizontal displacement d. y = v oy t + ½ at m = ½ (9.8 m/s 2 )t m = 4.9t 2 t = 0.49 s d = vt d = (0.42 m/s)(0.49 s) d = 0.21 m

20 In another experiment on the same table, the target ball B is replaced by target ball C of mass 0.10 kg. The incident ball A again slides at 1.4 m/s, as shown above left, but this time makes a glancing collision with the target ball C that is at rest at the edge of the table. The target ball C strikes the floor at point P, which is at a horizontal displacement of 0.15 m from the point of the collision, and at a horizontal angle of 30° from the +x ‑ axis, as shown above right.

21 c.Calculate the speed v of the target ball C immediately after the collision. v c = d/t v c = 0.15 m/0.49 s v c = 0.31 m/s d.Calculate the y ‑ component of incident ball A's momentum immediately after the collision. p Ay = p cy p Ay = m c v c sin30° p Ay = (0.1 kg)(0.31 m/s)sin30° p Ay = kg·m/s


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