Physics 2514Lecture 26

P. Gutierrez

Department of Physics & AstronomyUniversity of Oklahoma

Physics 2514 – p. 1/10

Review

We have defined the following using Newton’s second law ofmotion (~Fnet = d~p

dt):

Impulse ~J =∫ tf

ti

~Fnet dt;Momentum ~p = m~v;Showed the impulse and momentum related by ~J = ∆~p.

Using Newton’s third law we arrived at momentumconservation (~F12 = −~F21 ⇒

d~Pnetdt

= 0):For two objects ~p1i + ~p2i = ~p1f + ~p2f ;Holds if no external forces acting on system;Holds for any number of isolated objects.Components independently conserved.

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Cases to Consider

Three basic problem typesUse impulse-momentum theorem to calculate changesin a systemCollisions

Perfectly inelastic—objects stick togetherPerfectly elastic—objects bounce apart (requiresenergy to be conserved)Inelastic—part way between the above two cases.

ExplosionsObjects initially at rest relative to each other

Collisions in 2 dimensionsIncludes cases given above.

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Example Impulse

A sled slides along on a horizontal surface on which thecoefficient of kinetic friction is 0.25. Its velocity at point A is8.0 m/s and at point B is 5.0 m/s. Use the impulse-momentumrelation to find how long the sled takes to travel from A to B.

An object is being slowed by friction (µk = 0.25). How long (∆t)does it take the object to reduce its speed from vi = 8.0 m/s tovf = 5.0 m/s?

MomentumPSfrag replacements

pi pf

A B

Forces

PSfrag replacements fk

mg

n

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Example Impulse

An object is being slowed by friction (µk = 0.25). How long (∆t)does it take the object to reduce its speed from vi = 8.0 m/s tovf = 5.0 m/s?

MomentumPSfrag replacements

pi pf

A B

Forces

PSfrag replacements fk

mg

n

J = F∆t = ∆p ⇒ − fk∆t = mvf − mvi

fk = nµk = mgµk ⇒ − mgµk∆t = m(vf − vi)

⇒ ∆t =vi − vf

gµk

= 1.2 s

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Clicker

A block of mass m is placed on a frictionless inclined plane thatmakes a 30◦ angle relative to the horizontal. The block isreleased from rest, travels a distance of 20 cm along the incline.At the bottom of the incline is a spring that applies a force givenby F = −100∆s, where ∆s is the amount the spring iscompressed. After striking the spring, the block starts movingupward and eventually comes to rest 10 cm below where itstarted. What is the total impulse from the moment the block isreleased until it comes to rest again.

A) What?B) 0 N-sC) 30 N-s

D) 100 N-sE) 200 N-s

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Example Momentum Conservation

Two ice skaters, with masses of 50 kg and 70 kg, are at thecenter of a 60 m diameter circular rink. The skaters push offeach other and glide to the opposite edges of the rink. If theheavier skater reaches the edge in 20 s, how long does it takethe lighter skater to reach the edge?

Objects A (m = 50 kg) and B (m = 70 kg) have a combined zeronet momentum. If object B takes 20 s to travel 30 m, how longdoes it take object A?

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Example Momentum Conservation

Objects A (m = 50 kg) and B (m = 70 kg) have a combined zeronet momentum. If object B takes 20 s to travel 30 m, how longdoes it take object A?

PSfrag replacements Before

After

p = 0

pA pB

x

Velocity object B:x = vBt ⇒ vB =

x

t= 1.5 m/s

Velocity object A:

mBvB − mAvA = 0

⇒ vA =mB

mA

vB = 2.1 m/s

Time for A to reach edge of rink:x = vAt ⇒ t =

x

vA

= 14.3 sPhysics 2514 – p. 8/10

Clicker

Consider a perfectly inelastic collision between two objects (acollision where the objects stick together after they interact).Object A has a mass m and an initial speed v, while object Bhas a mass 2m and is initially at rest. After they collide what isthe speed of the system

A) v

B) v/2

C) 2v

D) v/3

E) 3v

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Example 2-d Collision

The figure shows a collision between 3 clay objects. What is thespeed and direction of the resulting object?

BeforeAfter

PSfrag replacements

m = 90 g

p = 90~vf

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Example 2-d Collision

The figure shows a collision between 3 clay objects. What is thespeed and direction of the resulting object?

Momentum conservation, vertical

m20v2 − m40v4 sin(45◦) = m90vyf

⇒ vyf = −0.81 m/s

Momentum conservation, horizontal

m40v4 cos(45◦) − m30v3 = m90vxf

⇒ vxf = 0.26 m/s

Resultv =

q

v2

xf+ v2

yf= 0.85 m/s θ = tan−1

vyf

vxf

= −72.4◦

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Assignment

Start reading Chapter 10

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