006 Momentum Lab PDF

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Introductory Physics Hunter College Lab #6 6/17/21 Momentum Talha Faisal Dr. Robert Marx Summer 2021

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Objectives: You will explore concepts of momentum, conservation of momentum, conservation of kinetic energy for elastic and inelastic collisions in one dimension. Momentum and Its Relation to Force Momentum is a vector symbolized by the symbol p, and is defined as p mv eq. 1

The rate of change of momentum is equal to the net force:  p F  t eq. 2

The total momentum of an isolated system of objects remains constant . This is the Law of Conservation of Momentum Δp = 0, which means

eq. 3 when

eq. 4

Collisions and Impulse During a collision, objects are deformed due to the large forces involved. Since the force is equal to the change in momentum divided by time, we can write:

F t = p eq. 5 The Impulse is defined as:

Impulse F t eq. 6 The impulse tells us that we can get the same change in momentum with a large force acting for a short time, or a small force acting for a longer time.

Conservation of Energy and Momentum in Collisions Which types of collisions conserve KE or not Page 2 of 13

Here we have two objects colliding elastically. We know the masses and the initial speeds, before the collision, vA and vB

Before the Collision

After the Collision

total KE before total KE after 1 2

mAvA2  12 mBvB2  21 mAvA2  21 mBvB2

[elastic collison] eq. 7

include cons. of mom equation Since both momentum and kinetic energy are conserved, we can write two equations. This allows us to solve for the two unknown final speeds v’A and v’B.

vA  vB vB  vA or vA  vB  (vA  vB )

[head-on (1-D) elastic collision]

eq. 8

Pre-Lab Questions 1. A truck going has a head-on collision with a small car going what?? Which statement best describes the situation? Page 3 of 13

(a) The truck has the greater change of momentum because it has the greater mass. (b)The car has the greater change of momentum because it has the greater speed. (c) Neither the car nor the truck changes its momentum in the collision because momentum is conserved. (d)They both have the same change in magnitude of momentum because momentum is conserved. (e) None of the above is necessarily true. Explanation: There is no external force on the system and the change in linear momentum is zero meaning that the change in momentum is the same for both the car and the truck. 2. A small boat coasts at constant speed under a bridge. A heavy sack of sand is dropped from the bridge onto the boat. The speed of the boat (a) increases. (b)decreases. (c) does not change. (d)Without knowing the mass of the boat and the sand, we can’t tell. Explanation: There is no external force acting on the boat and the sand system, the total linear momentum of the system is going to be conserved. The final velocity is going to end up less than the initial velocity. 3. Two identical billiard balls traveling at the same speed have a head-on collision and rebound. If the balls had twice the mass, but maintained the same size and speed, how would the rebound be different? (a) At a higher speed. (b)At slower speed. (c) No difference. Explanation: Momentum of a system is conserved when the external force acting upon it is equal to zero, since both billiard balls have the same mass and the same speed the final momentum acter collision is equal. Doubling the mass would not have an effect on the final velocity.

4. An astronaut is a short distance away from her space station without a tether rope. She has a large wrench. What should she do with the wrench to move toward the space station? (a) Throw it directly away from the space station. Page 4 of 13

(b)Throw it directly toward the space station. (c) Throw it toward the station without letting go of it. (d)Throw it parallel to the direction of the station’s orbit. (e) Throw it opposite to the direction of the station’s orbit. Explanation: Throwing the wrench away from the space station will mean that the astronaut will be going to be moving towards the space station. The force she exerts on the wrench will be equal and opposite to the force that the wrench exerts on her, so throwing it away from the station will push her towards the station. Procedures Momentum 5. Go to http://physics.bu.edu/~duffy/HTML5/collisions_1D.html 6. Explore the different tabs that control the red and blue carts. Impulse - You will relate impulse to Newton’s Third Law of Motion. Select the red cart, Cart 1 to have an initial velocity of 1m/s and the blue cart, Cart 2, to have an initial velocity of -1m/s and select the elasticity to 1.0 (elastic collision). Make sure the Momentum graph tab is selected, so you see the Momentum vs. time graph displayed:

7. Run the simulation by clicking the Play tab. 8. Think about the impulse for each cart in terms of force on each cart. This is a force pair interaction (action-reaction pair). Do you expect the magnitudes of the force on each cart to be the same? Explain. If both carts 1 and 2 have the same mass, the magnitude of the force on each cart should be the same. Both carts are colliding at the same speed the magnitude of their forces must be the same. 9. Now, calculate the change in momentum, which is also the impulse, for each cart. Show the calculation. ΔP/ Δt = M Cart 1: (-1 m/s – 1 m/s) / ( 3s -0) = -0.67 m/s Cart 2: (1 m/s-(-1 m/s))/( 3s - 0) = 0.67 m/s

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10. How does impulse of the cart relate to Newton’s Third Law of Motion, given the force happens at the same time, for each cart? Since the amplitude of the impulse is the same for both carts. If all carts are subjected to the same force, then they will have the same impulse. 11. Assume the mass is 10kg. If the force 200 N, how long is the collision? m= 10 kg F = 200 N t= ? v = 1 m/s P = (10 kg)( 1m/s) = 10 kg * m/s 200 N = 10 / t t = 0.05 s Inelastic and Elastic Collisions – You will explore the conservation of momentum and kinetic energy. 12. Now, set the initial velocity of Cart 2 to zero. Keep all other parameters the same. Run the simulation.

13. Calculate the final velocities for Cart 1 and Cart 2, enter your results in Table 1. 14. Now run the simulations for the other 2 initial conditions listed in Table 1. 15. Calculate the final velocities for the 2 runs, enter your results in Table1.

Table 1. Elastic Collision Cart 1 mass

Cart 2 mass

Cart 1

Cart 2

Cart 1

Cart 2 Page 6 of 13

Initial Initial Final Velocity Velocity Velocity m

m

1 m/s

0 m/s

0 m/s

3m

m

1 m/s

0 m/s

0.50 m/s

3m

m

1 m/s

-1 m/s

0 m/s

Final Velocity 1 m/s

1.50 m/s

2 m/s

16. Show that the kinetic energy is conserved for the last simulation. Show your work 1/2 m1(v1)2 + 1/2 m2(v2)2 = 1/2 m1(v1)2 + 1/2 m2(v2)2 1/2 (3(1)2) + 1/2(1(-1)2) = 1/2 (3(0)2) + (1/2 (1(2)2) 2J = 2J Kinetic energy is conserved. 17. Now, set the “Elasticity of collision” to 0, which is completely inelastic. 18. Run the simulations when the carts have the same mass m, with Cart 1’s initial velocity set at 1 m/s, while Cart 2 is at rest. 19. Calculate the final velocity for Cart 1 and Cart 2, enter your results in Table 2. 20. Calculate the Kinetic Energy Lost for the collision, enter your result in Table 2. 21. Repeat step 18 – 20 for the other two cart conditions in Table 2. Table 2. Inelastic Collision Cart 1 mass

Cart 2 mass

Cart 1

Cart 2

m

m

1 m/s

0 m/s

3m

m

1 m/s

0 m/s

3m

m

1 m/s

-1 m/s

Initial Initial Velocity Velocity

Combine Kinetic Energy d Final Velocity Lost

0.50 m/s 0.75 m/s 0.50

2.5 J 3.75 J -5 J Page 7 of 13

m/s

22.

What the “hell” happened to the Kinetic Energy Lost? Explain.

The kinetic energy was never lost, it was converted to another form of energy like mechanical energy or thermal energy. If energy was actually lost, then that would go against the law of conservation of energy.

Calculation Kinetic Energy Lost: Before = After (1/2)(mv^2) + (½)m2v2^2 = (½)mtvt^2 + lost (½)(10)(1^2)+(½)(10)(0^2)= (½)(10+10)(.5^2) + lost 5 = 2.5+lost lost= 2.5J

(½)(mv^2) + (½)m2v2^2 = (½)mtvt^2 + lost (½)(30)(1^2)+(½)(10)(0^2)= (½)(30+10)(.75^2) + lost 15 = 11.25+lost lost= 3.75J

(½)(mv^2) + (½)m2v2^2 = (½)mtvt^2 + lost (½)(30)(1^2)+(½)(10)(-1^2)= (½)(30+10)(.5^2) + lost 0 = 5+lost lost= -5J

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A Ballistic Pendulum You will explore the momentum and mechanical energy via the Ballistic Pendulum A ballistic pendulum is a device that could be used by a crime-scene investigator to determine the muzzle velocity of a gun. The bullet comes out of the gun at high speed, and embeds itself into a target that is hanging from a string. This causes the target (with the bullet embedded) to swing back and forth, pendulum style. By knowing the maximum height that the target initially reaches, the speed of the bullet can be determined. Think about the Ballistic Pendulum, see figure below,

23. Write the equation for the conservation of momentum of the moving bullet (red dot on the left) and the stationary bob of the pendulum, just before and just after the bullet collides with the bob. mtgh=½*mt(v’)^2 and m1v1+m2v2=(m1+m2)*Vf. 24. Write the equation of the mechanical energy of the bullet-bob system once the bullet is embedded in the bob and the highest point they reach. ½ (m+M)vf2 = (m+M)gh 25. Calculate the angle that the pendulum will make at its highest point (from the vertical axis); mass of the bullet is 0.03 kg the mass of the target hanging bob is 0.870 kg, the initial speed of the bullet is 100.0 m/s, and the length of the bob is 2.0m. Show your calculations. Page 9 of 13

v2 / 2g = (3.33)2 / (2 (9.8)) = = 0.5657b = 2 (1- cos(θ)) Θ = 44.20

26. Now, go to the Ballistic Pendulum Simulation http://physics.bu.edu/~duffy/HTML5/ballistic_pendulum.html 27. Run the simulation for a 30g bullet of muzzle velocity 1.00m/s, that embeds itself into 870g target; set the length of the pendulum to 2.0m. 28. Determine the simulation’s highest angle for the pendulum; use the Pause, the step>> and...