Physics 221Chapter 9

On a Collision Course ...

• Linear Momentum = mass x velocity

p = mv

• Conservation of Momentum: In a collision, momentum BEFORE the collision equals the momentum AFTER the

collision

Problem 1 . . . The Odd Couple I

• A model railroad car (mass = 100 g) collides and locks onto a similar stationary car. The coupled cars move as a unit on a straight and frictionless track. The speed of the moving cars is most nearly

• A. 1 m/s• B. 2 m/s• C. 3 m/s• D. 4 m/s

Solution 1 . . . The Odd Couple I

• Momentum BEFORE = Momentum AFTER• 100 x 4 +0 = 200 x v• v = 2 m/s

Problem 2 . . . The Odd Couple II

• Is the K.E. BEFORE = K.E. AFTER? In other words, is the K.E. conserved in this collision?

Solution 2 . . . The Odd Couple II

• K.E. BEFORE the collision = (1/2)(0.1)(16) +0 = 0.8J

• K.E. AFTER the collision = (1/2)(0.2)(4) +0 = 0.4J

• Half of the K.E. has mysteriously disappeared!

• BUT this does not mean that ENERGY is not conserved! K.E. was transformed (converted) into other forms of energy.

Problem 3 . . . Big Bang!

• A shell explodes into two unequal fragments. One fragment has twice the mass of the other. The smaller fragment moves in the NE direction with a speed of 6 m/s. The velocity (speed and direction) of the other fragment is most nearly

• A. 3 m/s SW• B. 3 m/s NE• C. 4 m/s NE• D. 4 m/s SW

Solution 3 . . . Big Bang!

• M x 6 = 2 M x V• V = 3 m/s in the SW direction.

Close collisions of the second kind ...

• There are TWO types of collisions:

• I. Momentum conserved and K.E. also conserved. This type is called an ELASTIC collision. Elastic collisions are “non-sticky” as in billiard balls or steel ball-bearings.

• II. Momentum conserved BUT K.E. NOT conserved. This type is called an INELASTIC collision. Inelastic collisions are “sticky” as in coupled railroad cars and putty.

Problem 4 . . . Elastic I

• Two hockey pucks collide elastically on ice. P2 is at rest and P1 strikes it “head-on” with a speed of 3 m/s.

• A. P1 stops and and P2 moves forward at 3 m/s• B. P1 and P2 move forward at 1.5 m/s• C. P1 and P2 move in opposite directions at 1.5 m/s• D. P1 moves forward at 1 m/s and P2 moves forward at

2 m/s

Solution 4 . . . Elastic I

• A. P1 stops and P2 moves forward at 3 m/s.

• Please verify that:• 1. Momentum is conserved• 2. K.E. is also conserved (elastic)

Problem5 . . . Elastic II

• Two hockey pucks collide elastically on ice. P2 has twice the mass of P1 and is at rest and P1 strikes it “head-on” with a speed of 3 m/s. Calculate their velocities after the collision.

Solution 5 . . . Elastic II

• P1 moves backward at 1 m/s and P2 moves forward at 2 m/s.

• HINT: • 1. Take 5 sheets of scrap paper ( Trust me!!!)• 2. Write the equation for momentum conservation• 3. Write the equation for K.E. conservation• 4. Solve two equations for two unknowns.

Tell me more about momentum!

• O.K. good boys and girls. Here is everything you always wanted to know about momentum but were afraid to ask!

• In the beginning there was F = ma• So F = m(vf - vi) / t

• If F = 0 then : mvf = mvi

• Another interesting observation: If F is NOT zero then momentum WILL change and change in momentum equals F x t. This is called IMPULSE

Problem 6 . . . Tiger in the woods

• A golf ball has a mass of 48 g. The force exerted by the club vs. time is a sharp spike that peaks at 180 N for 0.01 s and then drops back to zero as the ball leaves the club head at high speed. It is estimated that the average force is 90 N over a short time (t) of 0.04 s. The speed of the ball is

• A. 35 m/s• B. 75 m/s• C. 95 m/s• D. 135 m/s

Solution 6 . . . Impulse to ride the tiger

• Impulse = F t = 90 x 0.04 = 3.6 Ns• Impulse = change in momentum• 3.6 = mv -0• 3.6 = 0.048 x v• v = 75 m/s

Problem 7 . . . Teeter- Totter

• FB weighs 180 pounds and sits at the 20 cm. mark. Where should SP (120 pounds) sit in order to balance the teeter-totter at the playground?

0 20 50 100

180 120

Solution 7 . . . See-Saw

• 30 x 180 = what x 120• what = 45 or 95 cm. mark

0 20 50 100

180 120

Problem 8 . . . Center of Mass

• The C.M. is a weighted average position of a distribution of masses where the system can be balanced.

• Where is the C.M. ?

100300

0 20 50 80 100?

Solution 8 . . . Center of Mass

(100)( x -20) = (300)(80 - x)x = (100) (20) + (300)( 80) / (100 + 300)x = 65

0 20 50 80 100

100300

x

A formula for C.M.

C.M. = M1 * X1 + M2 * X2

M1 + M2

0 X1 50 X2 100

M 1M 2

C.M..

Problem 9 . . . more C. M.

• Include the mass of the meter stick (60 g). Where is the new C.M. ?

100300

0 20 50 80 100?

Solution 9 . . . more C. M.

C.M. = M1 * X1 + M2 * X2 + M3 * X3

M1 + M2 + M3

C.M. = (100)(20) + (60)(50) +(300)(80)

100 + 60 + 300

C.M. = 63

Problem 10 . . . The Hole

• A circular hole of radius R/2 is drilled in a disc of mass M and radius R as shown in the figure. The center of mass is closest to

• A. (R/6 , 0)• B. (R/4 , 0)• C. (R/3 , 0)• D. (R/2 , 0)

(R , 0)

Circular plate with hole

• If the mass of the missing piece is 100 then the mass of the remaining piece is 300 for a total of 400. In other words, if M is the mass of the original disc (no hole) then the mass of the disc with the hole is 3M/4 and the mass of the circular piece removed is M/4.

• NOTE: Non-Linear thinking! Disc with HALF the radius has only a quarter of the mass!

Solution 10A model for the “hole” problem

• (100)(R/2) = (300)(x)• x = R/6

100 300

x