# Chapter 6 – Linear Momentum and Collisions

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Chapter 6 – Linear Momentum and Collisions
Compare the inertia of a semi truck and a roller skate… Will the truck always have more inertia than the roller skate? Will it always have more momentum than the roller skate?

Warm-Up Estimate the amount of work the engine performed on a 1200-kg car as it accelerated at 1.2 m/s2 over a 150-meter distance.

Warm-Up Solution m = 1200 kg a = 1.2 m/s2 d = 150 m F = ma W = ∆KE=Fcosqd W = ma×d W = (1200 kg)(1.2 m⁄s^2 )(cos0o)(150 m) W = 2.2x10^5 J

A gallon of gasoline contains about 1. 3 x 108 joules of energy
A gallon of gasoline contains about 1.3 x 108 joules of energy. A kg car traveling at 20 m/s skids to a stop. Estimate how much gasoline it will take to bring the car back to the original speed. To complicate matters further, consider the fact that only about 15% of the energy extracted from gasoline actually propels the car. The rest gets exhausted as heat and unburnt fuel.

Linear Momentum: Inertia in Motion
Linear momentum of an object is equal to the product of its mass and velocity SI Units of momentum are kgm/s A lowercase p is used for the momentum of an individual particle

Total Linear Momentum of a System
The total linear momentum of a system is the vector sum of the momenta of the individual particles The total linear momentum of a system, expressed by an uppercase , is used for the total momentum of the individual particles of the system

Sum of Linear Momentum

pcar=mcarvcar= (1500 kg)(25 m/s) = 3.75 x 104 kgm/s
Example 1: Which has more momentum: a) a 1500 kg car moving at 25 m/s or b) a kg truck moving at 1.00 m/s? Given: mcar = 1500 kg vcar = 25 m/s mtruck = kg vtruck = 1 m/s pcar=mcarvcar= (1500 kg)(25 m/s) = 3.75 x 104 kgm/s b) ptruck = mtruckvtruck=(40000kg)(1m/s) =4.00 x 104 kgm/s The much slower truck has more momentum than the faster car because the truck has much greater mass.

Example 2: Two identical 1500 kg cars are moving perpendicular to each other. One is moving with a speed of 25.0 m/s due north, and the other is moving at 15.0 m/s due east. What is the total linear momentum of the two car system? Given: m1 = m2 = 1500 kg v1 = 25.0 m/s v2 = 15.0 m/s p1 = m1 v1 = (1500kg)(25.0m/s) = 3.75 x 104kgm/s p2 = m2v2 = (1500kg)(15.0m/s) = 2.25 x 104kgm/s Continued on next slide…

From vector addition, the magnitude of the total momentum is
And the direction is given by

1. Determine the momentum of a ... 60-kg halfback moving eastward at 9 m/s. 1000-kg car moving northward at 20 m/s. 40-kg freshman moving southward at 2 m/s. 2. A car possesses units of momentum. What would be the car's new momentum if ... its velocity were doubled. its velocity were tripled. its mass were doubled (by adding more passengers and a greater load) both its velocity were doubled and its mass were doubled.

Solutions 1. a) p = m*v = 60 kg*9 m/s p = 540 kg•m/s, east b) p = m*v = 1000 kg*20 m/s p = kg•m/s, north c) p = m*v = 40 kg*2 m/s p = 80 kg•m/s, south

Solutions 2. a) p = units (doubling the velocity will double the momentum) b) p = units (tripling the velocity will triple the momentum) c) p = units (doubling the mass will double the momentum) d) p = units (doubling the velocity will double the momentum and doubling the mass will also double the momentum; the combined result is that the momentum is doubled twice -quadrupled)

Newton’s Laws and Momentum
Newton’s second law, F=ma, can also be expressed in terms of momentum. If the mass is constant, then If an object’s momentum changes, then a force must have acted upon it.

Solve the problems on “Chapter 4” Momentum and Energy”
Practice Solve the problems on “Chapter 4” Momentum and Energy” Momentum p. 27 Worksheet

A 100. -kg crate is sliding down an plane inclined at an angle of 30
A 100.-kg crate is sliding down an plane inclined at an angle of 30. degrees. The coefficient of friction between the crate and the incline is Determine the net force and acceleration of the crate.

Impulse and Momentum Impulse is the change in momentum.
When a moving object stops, its impulse depends only on its change in momentum. This can be accomplished by a large force acting for a short time, or a smaller force acting for a longer time.

Impulse and the Bouncing Ball Experiment
Predict what will happen when two identical balls are dropped from the same height onto the same type of surface

“Chapter 4” Momentum and Energy”
Practice Solve the problems on “Chapter 4” Momentum and Energy” Impulse-Momentum p.31 Use the Rally Coach Method to work through the Examples 3, 4, 5, and 6.

Example 3 A hockey player applies an average force of 80.0 N to a 0.25 kg hockey puck for a time of 0.10 seconds. Determine the impulse experienced by the hockey puck.

Solution to Example 3: A hockey player applies an average force of 80
Solution to Example 3: A hockey player applies an average force of 80.0 N to a 0.25 kg hockey puck for a time of 0.10 seconds. Determine the impulse experienced by the hockey puck. Given: F=80.0 N m=0.25 kg t=0.10 s I = FDt

Example 4: When bunting, a baseball player uses the bat to change both the speed and direction of the baseball. (a) How will the magnitude of the momentum of the baseball before and after the bunt change? (Describe in words, not numbers) (b) The baseball has a mass of 0.16 kg; its speeds before and after the bunt are 15m/s and 10 m/s respectively; the bunt lasts s. What is the change in momentum of the baseball? (c) What is the average force on the ball by the bat?

Solution to Example 4 (b) Choose the direction of motion before the bunt as positive. v = 10 m/s, vo = 15 m/s. p = mv – mvo = (0.16 kg)(10 m/s) – (0.16 kg)(15 m/s) =  4.0 kgm/s in direction opposite v0 (c)

Example 5 A boy catches—with bare hands and his arms rigidly extended—a 0.16-kg baseball coming directly toward him at a speed of 25 m/s. He emits an audible “Ouch!” because the ball stings his hands. He learns quickly to move his hands with the ball as he catches it. If the contact time of the collision is increased from 3.5 ms to 8.5 ms in this way, how do the magnitudes of the average impulse forces compare?

Solution to Example 5 Given: m=0.16 kg, v0=25 m/s, v=0m/s, t1=3.5ms, t2=8.5ms Favg t = mv  mvo = mvo,  So the magnitude is

Example 6 A karate student tries not to follow through in order to break a board. How can the abrupt stop of the hand (with no follow-through) generate so much force?   Assume that the hand has a mass of 0.35 kg and that the speeds of the hand just before and just after hitting the board are and 0, respectively. What is the average force exerted by the fist on the board if (a) the fist follows through, so the contact time is 3.0 ms, and (b) the fist stops abruptly, so the contact time is only 0.30 ms?

Solution to Example 6 By stopping, the contact time is short. From the impulse momentum theorem (Favg t = p = mv  mvo), a shorter contact time will result in a greater force if all other factors (m, vo, v) remain the same.