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Topic 2.2 Extended D – Impulse. Newton's Second Law ( p -form). Impulse. Topic 2.2 Extended D – 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.
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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 = Ft We call the quantity "Ft" 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 = Ft
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.
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.
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.
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.
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.
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
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.
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
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 kgm/s pf = +50(0.14) x After pf = +7 x kgm/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
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 kgm/s p0y = 0 j kgm/s pfx = +50cos30°(0.14) i After vf = +50 i m/s pfx = +6.062 i kgm/s pfy = +50sin30°(0.14) j pfx = +3.5 j kgm/s
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 kgm/s pfy = +3.5 j kgm/s p0x = -5.6 i kgm/s p0y = 0 j kgm/s ∆py = (+3.5 - 0) j ∆px = (+6.062--5.6) i ∆px = 11.662 i kgm/s ∆py = 3.5 j kgm/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
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 = Ft J = Ft J =Area under F vs. t graph