Topic 2.2 ExtendedD – Impulse

Recall the p-form of Newton's 2nd law:

We can rewrite this equation so that p is isolated:

Topic 2.2 ExtendedD – Impulse

F = pt

Newton's Second Law (p-form)

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 Impulse

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.

Note how P appears to be conserved.

Collision between an alpha particle and a nucleus.

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.

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

Collision between a knife and a water balloon. Note how the water retains its shape for an instant, satisfying Newton’s 1st law.

A collision is an event in which a relatively strong force acts on two or more bodies for a relatively short time.

Topic 2.2 ExtendedD – Impulse

Consider two pool balls that collide:

“Before” phase

“During” phase

“After” phase

FYI: We define the following three phases of a collision.

system boundary

system boundary

system boundary

FYI: Since there are no external forces in the x-direction, ∆Px = 0.

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:

A B

A B

A B

A B

A B

vAi

vAf

vBi

vBf

“Before” phase

“During” phase

“After” phase

t

F

Bef

ore

During

Aft

er

FYI: Note the perfect symmetry because of the action-reaction force pair. Note also that the “During” phase is where all the action happens.

FYI: Don’t forget that even though the forces are identical (except for direction), they are ACTING ON DIFFERENT BODIES.

FAB FBA

FAB FBA

FAB FBA

Topic 2.2 ExtendedD – Impulse

we can define the average force F as follows:Although the force varies with time,

t

-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):

F

-The area of the rectangle is the same as the area of the original impulse curve.

F

∆t

J = Area under F vs. t graph

J = 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

We sketch the before and after pictures here:

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.

v0 = -40 x m/s

vf = +50 x m/s

Before

After

p0 = -40(0.14) xp0 = -5.6 x kgm/s

pf = +50(0.14) xpf = +7 x kgm/s

J = ∆p = pf – p0 = +7 x – -5.6 x = +12.6 x

J = < F >∆t ⇒ 12.6 x = < F > 0.0012⇒ < F > = 10500 n

FYI: Since a newton is about a quarter pound, <F> is about 10500/4 = 2626 pounds (more than a ton).•Furthermore, the maximum force is greater than the average force!

Fmax<F>

Topic 2.2 ExtendedD – Impulse

We sketch the before and after pictures here:

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.

v0 = -40 i m/s

vf = +50 i m/s

Before

After

p0 = -40(0.14) ip0x = -5.6 i kgm/s

pfx = +50 cos 30°(0.14) i

pfx = +6.062 i kgm/s30°

p0y = 0 j kgm/s

pfy = +50 sin 30°(0.14) j

pfx = +3.5 j kgm/s

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 = -5.6 i kgm/s

∆px = (+6.062--5.6) i

p0x = +6.062 i kgm/s

p0y = 0 j kgm/s

pfy = +3.5 j kgm/s

∆px = 11.662 i kgm/s

∆py = (+3.5 - 0) j

∆py = 3.5 j kgm/s

Jx = ∆px= 11.662 i Jy = ∆py

= 3.5 j

Jx = < Fx >∆t Jy = < Fy >∆t

11.662 i = < Fx >(.0012)

< Fx > = 9718.33 i n

3.5 j = < Fy >(.0012)

< Fy > = 2916.67 j n

FYI: The magnitude of the force is given by <F>2 = Fx2 + Fy

2 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

Since momentum and kinetic energy both have m's and v's in them, we can find a relationship between the two:

K = mv212

p = mv

v = pm

v2 = p2

m2

K = m12

p2

m2

K = p2

2mKinetic Energy and Momentum

A summary:

p = Ft

J = Ft

J = Area under F vs. t graph

p = J Equations of Impulse