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
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