1. Momentum and Collisions

2. Momentum The linear momentum of an object of mass m moving with a velocity v is the product of the mass and the velocity. The linear momentum of an object of mass m moving with a velocity v is the product of the mass and the velocity. Momentum = mass X velocity Momentum = mass X velocity p=mv p=mv Units kg ● m/s Units kg ● m/s

3. Momentum Practice A 2250 kg pickup truck has a velocity of 25 m/s to the east. What is the momentum of the truck? A 2250 kg pickup truck has a velocity of 25 m/s to the east. What is the momentum of the truck? Given: m=2250 kg Given: m=2250 kg v= 25 m/s v= 25 m/s p=mv 2250kg x 25 m/s= p=mv 2250kg x 25 m/s= 5.6x10 4 kg ● m/s 5.6x10 4 kg ● m/s

4. Changing Momentum A change in momentum takes force and time A change in momentum takes force and time F=m(Δv/Δt) F=m(Δv/Δt) Newtons original formula Newtons original formula F=Δp/Δt F=Δp/Δt Δp=mv f -mv i Δp=mv f -mv i Force= change in momentum during time interval Force= change in momentum during time interval

5. Impulse Impulse is the force for the time interval. Impulse is the force for the time interval. Impulse-momentum theorem is the expression FΔt= Δp Impulse-momentum theorem is the expression FΔt= Δp If you extend the time of impact you reduce the amount of force. If you extend the time of impact you reduce the amount of force.

6. Force and Impulse A 1400 kg car moving westward with a velocity of 15 m/s collides with a utility pole and is brought to rest in.30 s. Find the force exerted on the car during the collision. A 1400 kg car moving westward with a velocity of 15 m/s collides with a utility pole and is brought to rest in.30 s. Find the force exerted on the car during the collision. Given m=1400 kg, Δt=.30 s Given m=1400 kg, Δt=.30 s v f =0 m/s v i =15 m/s v f =0 m/s v i =15 m/s What formula can I use? What formula can I use?

7. Impulse-Momentum FΔt= Δp FΔt= Δp Δp=mv f -mv i Δp=mv f -mv i FΔt=mv f -mv I am looking for force FΔt=mv f -mv I am looking for force F=mv f -mv/ Δt Now I just plug in my numbers F=mv f -mv/ Δt Now I just plug in my numbers (1400*0-1400*-15)/.30= 7.0*10 4 N to the east (1400*0-1400*-15)/.30= 7.0*10 4 N to the east

8. Stopping Distance A 2240 kg car traveling west slows down uniformly from 20.0 m/s to 5.0 m/s How long does it take the car to decelerate if the force on the car is 8410 to the east? How far does the car travel during the deceleration? A 2240 kg car traveling west slows down uniformly from 20.0 m/s to 5.0 m/s How long does it take the car to decelerate if the force on the car is 8410 to the east? How far does the car travel during the deceleration?

9. What am I given? M= 2240 kg M= 2240 kg v i = 20.0 m/s to the west = -20.0 m/s v i = 20.0 m/s to the west = -20.0 m/s v f = 5.0 m/s to the west = -5.0 m/s v f = 5.0 m/s to the west = -5.0 m/s F= 8410 N to the east so it stays positive F= 8410 N to the east so it stays positive Unknown Δt and Δx Unknown Δt and Δx

10. Impulse Momentum Theorem FΔt= Δp FΔt= Δp Δp=mv f -mv i Δp=mv f -mv i Δt=mv f -mv i /F Δt=mv f -mv i /F ((2240*-5)-(2240*-20))/8410 ((2240*-5)-(2240*-20))/8410 Δt=4.0s Δt=4.0s

11. Displacement of the vehicle To find the change in distance To find the change in distance Average velocity= change in d/change in t Average velocity= change in d/change in t Average velocity =(initial +final)/2 Average velocity =(initial +final)/2 Since these are equal I can combine them Since these are equal I can combine them Δx/Δt=(v f + v i )/2 Δx/Δt=(v f + v i )/2 Solving for x=((v f + v i )/2) *Δt Solving for x=((v f + v i )/2) *Δt x=((-20+-5)/2)*4 x=((-20+-5)/2)*4 x=-50 so it would be 50m to the west x=-50 so it would be 50m to the west

12. Conservation Of Momentum When I have two objects A and B When I have two objects A and B p(A i )+p(B i )=p(A f )+p(B f ) p(A i )+p(B i )=p(A f )+p(B f ) Total initial momentum=total final momentum Total initial momentum=total final momentum

13. Conservation of Momentum A 76 kg boater, initially at rest in a stationary 45 kg boat, steps out of the boat and onto the dock. If the boater moves out of the boat with a velocity of 2.5 m/s to the right what is the final velocity of the boat? A 76 kg boater, initially at rest in a stationary 45 kg boat, steps out of the boat and onto the dock. If the boater moves out of the boat with a velocity of 2.5 m/s to the right what is the final velocity of the boat? Given: m of person= 76 kg v i =0 m/s,v f =2.5 to the right Given: m of person= 76 kg v i =0 m/s,v f =2.5 to the right m of boat= 45 kg v i =0 v f =? m of boat= 45 kg v i =0 v f =?

14. m 1 v 1i +m 2 v 2i =m 1 v 1f +m 2 v 2f Because the boater and the boat are initially at rest m 1 v 1i + m 2 v 2i =0 Rearrange to solve for final velocity of boat v 2f = (-m 1 /m 2 )/v 1f -76/45 * 2.5=4.2 to the right -4.2

15. Two train cars of mass 8000 kg are traveling toward each other. One at a speed of 30 m/s to the west and the other at a speed of 25 m/s to the east. They crash and become entangled, what is their final velocity and direction? Two train cars of mass 8000 kg are traveling toward each other. One at a speed of 30 m/s to the west and the other at a speed of 25 m/s to the east. They crash and become entangled, what is their final velocity and direction?

16. m m 1 =8000 kg m 2 = 8000 kg v 1 =-30m/s v 2 =25 m/s m 1 v 1 +m 2 v 2 =m tot v tot