Topic 2.2 Extended D – Impulse

1. Topic 2.2 ExtendedD – Impulse

2. Newton's Second Law (p-form) Impulse Topic 2.2 ExtendedD – Impulse Recall the p-form of Newton's 2nd law: p t F = We can rewrite this equation so that p is isolated: p = Ft We call the quantity "Ft" the impulse and denote it with the letter J. In words... The impulse imparted to an object is equal to the product of the average force acting on it, and the time over which that force acts. J = Ft

3. Topic 2.2 ExtendedD – Impulse A collision is an event in which a relatively strong force acts on two or more bodies for a relatively short time. The Meteor Crater in the state of Arizona was the first crater to be identified as an impact crater. Between 20,000 to 50,000 years ago, a small asteroid about 80 feet in diameter impacted the Earth and formed the crater.

4. Topic 2.2 ExtendedD – Impulse A collision is an event in which a relatively strong force acts on two or more bodies for a relatively short time. A Cosmic Collision Between Two Galaxies, UGC 06471 and UGC 06472. Although this type of collision is long-lived by our standards, it is short-lived as measured in the lifetime of a galaxy.

5. Topic 2.2 ExtendedD – Impulse A collision is an event in which a relatively strong force acts on two or more bodies for a relatively short time. Collision between analpha particleand anucleus. Note how P appears to be conserved.

6. Topic 2.2 ExtendedD – Impulse A collision is an event in which a relatively strong force acts on two or more bodies for a relatively short time. Collision between a bullet and an apple.

7. Topic 2.2 ExtendedD – Impulse A collision is an event in which a relatively strong force acts on two or more bodies for a relatively short time. Collision between a knife and a water balloon. Note how the water retains its shape for an instant, satisfying Newton’s 1st law.

8. Topic 2.2 ExtendedD – Impulse FYI: Since there are no external forces in the x-direction, ∆Px= 0. Consider two pool balls that collide: FYI: We define the following three phases of a collision. system boundary “Before” phase system boundary “During” phase system boundary “After” phase

9. vAi F B A vBf B A t B A B A vAf vBi B A FYI: Don’t forget that even though the forces are identical (except for direction), they are ACTING ON DIFFERENT BODIES. Topic 2.2 ExtendedD – Impulse The previous example is that of a single collision. If we take a close-up look at a collision between two bodies, we can plot the force vs. the time acting on each mass during the collision: “Before” phase FAB FBA FAB FBA “During” phase FAB FBA After Before During “After” phase FYI: Note the perfect symmetry because of the action-reaction force pair. Note also that the “During” phase is where all the action happens.

10. F ∆t F t impulse in terms of average force FYI: Sometimes the average of a value, say the average force, is denoted <F> rather than F. Topic 2.2 ExtendedD – Impulse Although the force varies with time, we can define the average force F as follows: -Imagine an ant farm (two sheets of glass with sand in between) filled (with sand) in the shape of the above impulse curve: -We now let the sand level itself out (by shaking the ant farm): -The area of the rectangle is the same as the area of the original impulse curve. J =Area under F vs. t graph J = F∆t

11. Fmax <F> FYI: Since a newton is about a quarter pound, <F> is about 10500/4 = 2626 pounds (more than a ton). Topic 2.2 ExtendedD – Impulse •Furthermore, the maximum force is greater than the average force! A 0.14-kg baseball comes in at 40 m/s, strikes the bat, and goes back out at 50 m/s. If the collision lasts 1.2 ms (a typical value), find the average force exerted on the ball by the bat during the collision. Assume horizontal velocities. We sketch the before and after pictures here: v0 = -40 x m/s p0 = -40(0.14) x Before p0 = -5.6 x kgm/s pf = +50(0.14) x After pf = +7 x kgm/s vf = +50 x m/s = pf – p0 J = ∆p = +7 x – -5.6 x = +12.6 x J =< F>∆t ⇒ 12.6 x =< F> 0.0012 ⇒ < F> = 10500 n

12. 30° Topic 2.2 ExtendedD – Impulse A 0.14-kg baseball comes in horizontally at 40 m/s, strikes the bat, and goes back out at 50 m/s at a 30° elevation. If the collision lasts 1.2 ms find the average force exerted on the ball by the bat during the collision. We sketch the before and after pictures here: v0 = -40 i m/s p0 = -40(0.14) i Before p0x = -5.6 i kgm/s p0y = 0 j kgm/s pfx = +50cos30°(0.14) i After vf = +50 i m/s pfx = +6.062 i kgm/s pfy = +50sin30°(0.14) j pfx = +3.5 j kgm/s

13. FYI: The magnitude of the force is given by <F>2 = Fx2 + Fy2 so that <F> = 10146.6 n. FYI: The elevation angle of the force is given by  = tan-1(Fy/Fx) so that  = 16.7° ( not the same as the final direction of the ball). Topic 2.2 ExtendedD – Impulse A 0.14-kg baseball comes in horizontally at 40 m/s, strikes the bat, and goes back out at 50 m/s at a 30° elevation. If the collision lasts 1.2 ms find the average force exerted on the ball by the bat during the collision. p0x = +6.062 i kgm/s pfy = +3.5 j kgm/s p0x = -5.6 i kgm/s p0y = 0 j kgm/s ∆py = (+3.5 - 0) j ∆px = (+6.062--5.6) i ∆px = 11.662 i kgm/s ∆py = 3.5 j kgm/s Jy =∆py =3.5 j Jx =∆px =11.662 i Jy =< Fy>∆t Jx =< Fx>∆t 11.662 i =< Fx>(.0012) 3.5 j =< Fy>(.0012) < Fx> = 9718.33 i n < Fy> = 2916.67 j n

14. Kinetic Energy and Momentum Equations of Impulse Topic 2.2 ExtendedD – Impulse p =mv Since momentum and kinetic energy both have m's and v's in them, we can find a relationship between the two: p m 1 2 v = K =mv2 p2 m2 v2 = p2 m2 1 2 K =m p2 2m K = A summary: p = J p = Ft J = Ft J =Area under F vs. t graph