IMPULSE-AND-MOMENTUM-PHYSICS-PRESENTATION.pptx

Impulse and Momentum Presentation

1. The Impulse & Momentum Lesson 6.2

2. Learning Objectives • For this lesson, we will know the definition of Impulse and Momentum. • In this lesson, you can learn how to solve regards with impulse and momentum. • We will discuss about the conservation of linear momentum.

3. Impulse-momentum theorem The impulse-momentum theorem, considered as an alternative statement of Newton 2nd law of motion, states that the rate of change of momentum of a body is equal to the force applied on it. (Formula: I=mv−mu or Ft= mv-mu )

4. The IMPULSE AND MOMENTUM Is a vector quantity , with the same direction as the direction of velocity. Momentum can be defined as "mass in motion" All objects have mass; so if an object is moving, then it has its mass in motion. Is the product of mass (m) of the object and its velocity (v). Momentum

5. The IMPULSE AND MOMENTUM Impulse is defined as the product force acting on an object and the time during which the force acts, which means impulse is the change in momentum . Impulse is a vector quantity and has the same direction as the average force.

6. SI unit: • Impulse is N•s • Momentum is kg•m/s

7. The Formula: • Impulse= F∆t • Momentum: a) p=mv-mu b) F∆t=mv-mu

8. Exercise Time! (Let's try to solve the exercise problems)

9. Problem Exercises Hypervelocity bullets used in 0.22 long rifle usually weigh around 2.1 g and can have a speed of 550 m/s. What must the speed of a 75 kg man to match the momentum of its bullet? 1.

10. Problem Exercises 2. A neophyte player catches a 125 g ball moving at 25.0 m/s in 0.02 s. A proffesional player catches the same ball in 1.0 s by slightly retracting his hand during the catch. Find the forces exerted by the ball on the hands of the two players.

11. CONSERVATION OF LINEAR MOMENTUM Conversion of Impulse Momentum: The principle of conservation of momentum states that if two objects collide, then the total momentum before and after the collision will be the same if there is no external force acting on the colliding objects.

12. CONSERVATION OF LINEAR MOMENTUM o Internal forces – the forces that the particles of a system exert on one another. o External forces – forces exerted on any part of the system by other objects outside the system. Internal and External Forces

13. Examples of Conservation of Linear Momentum A person standing on a skateboard. When the person pushes off the ground in one direction, the skateboard moves in the opposite direction, demonstrating the conservation of momentum. 1.

14. Examples of Conservation of Linear Momentum 2. A person stepping from a small boat onto a dock. As the person moves toward the dock, the boat moves away. Again, the total momentum of the system (person + boat) remains constant, but the distribution of momentum changes, leading to the observed motion.

15. Examples of Conservation of Linear Momentum 3. This is also applicable in rocket propulsion. When a rocket is fired, the expelled exhaust gases move downward at high speed. The downward momentum of the gases is equal and opposite to the upward momentum of the rocket, resulting in balanced forces.

16. Examples of Conservation of Linear Momentum 3. This upward force exerted by the gases is known as thrust. The relationship between force, time, and momentum change is described by the impulse-momentum theorem.

17. Exercise Time! (Let's try to solve the exercise problems)

18. A 50.0 kg student uses an improvised 75 kg raft in order to cross a heavily flooded street. He noticed that as he jumps to the sidewalk opposite the street with a speed of 1.5 m/s relative to the flood, the raft moves away. With what the speed will the raft move relative to flood? Assume that the raft is stationary before the student jumps to the sidewalk. Neglect the fluid friction. Exercise Problem (1)

19. Consider the raft and the student as an isolated system. Let subscripts R and S. The momentum of the system before the student jumps is equal to zero. Solution Formula: 0= msvs + mrvr -mrvr=msvs vr=-msvs/mr Solution: =(-50.0 kg)(1.5m/s)/ 75 kg =-1.0 m/s Formula Solution Conclusion: Since it is negative for velocity, therefore the raft moves backward. Conclusion

20. A child accidentally dropped a 0.250 kg plate to the floor. The plate broke into 3 pieces: a 0.125 kg piece moves in the +y direction with a speed of 1.2 m/s and a 0.053 kg piece in the +x direction at a speed of 1.6 m/s. Find (a) the mass of the third piece and (b) the horizontal and vertical components of its velocity. Exercise Problem (2)

21. SOLUTION:

22. SOLUTION: The momentum of the two pieces (p, and P2) are computed using Eq. (6.6). P1= (0.125 kg) (1.2 m/s) = 0.15 kg •m/s, along the +y-direction P2 = (0.053 kg) (1.6 m/s) = 0.0848 kg •m/s, along the +x-direction Table A. m3 = 0.250 kg-0.125 kg - 0.053 kg = 0.072 kg

23. SOLUTION: B. To determine the horizontal (v,*) and vertical (Vy) components of the velocity of the third piece, apply the conservation of momentum for the horizontal and vertical momenta. Since the plate has no horizontal motion as it falls on the floor, there is no horizontal momentum before the interaction. The conservation of horizontal momentum gives 0 = 0.085 kg •m/s + P3x P3x = - 0.085 kg • m/s. Therefore, V3x = -0.085 kg • m/s / 0.072kg = -1.181 m/s ~ -1.2 m/s. The conservation of vertical momentum gives 0=0.15 kg •m/s + p3y P3y = -0.15 kg• m/s V3y = -0.15 kg • m/s / 0.072 kg = -2.083 m/s ~ -2.1 m/s.A. m3 = 0.250 kg-0.125 kg - 0.053 kg = 0.072 kg

24. Question Time! What have you learned?

25. Thank You