physics ch. 6
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Physics ch. 6. Momentum and Collisions. Vocab Words. Elastic collision Impulse Momentum Perfectly inelastic collision. 6-1Momentum and Impulse. In this chapter, we will examine how the force and duration of a collision between two objects affects the motion of the objects - PowerPoint PPT PresentationTRANSCRIPT
Physics ch. 6
Momentum and Collisions
Vocab Words
Elastic collision Impulse Momentum Perfectly inelastic collision
6-1Momentum and Impulse
In this chapter, we will examine how the force and duration of a collision between two objects affects the motion of the objects
The linear momentum of an object is defined as the product of the mass and the velocity of the object.
P=mv
Momentum is a vector quantity, direction matches the direction of the velocity
More massive object has more momentum than a less massive object at the same velocity
However, small objects with a large velocity also have a large momentum
Ex: hail
6A Sample Problem
A 2250 kg pickup truck has a velocity of 25 m/s to the east. What is the momentum of the truck?
A change in momentum takes force and time
Ex: It takes more force to stop a fast moving ball than a slower moving ball
Ex: It takes more force to stop a real dump truck moving at the same speed at a toy dump truck
Momentum is closely related to force
F = Δp/Δt
Impulse-Momentum Theorem
FΔt = Δp or FΔp = mvf – mvi
Where FΔt is the impulse
A small force acting for a long time can produce the same change in momentum as a large force acting for a short time.
In this book, all forces exerted on an object are assumed to be constant unless otherwise stated.
This equation helps describe some common sports examples:
Ex: baseball-the ball will experience a greater momentum if the force of the bat is kept in contact with the ball for a longer time period (FΔt)
6B Sample Problem
A 1400 kg car moving westward with a velocity of 15 m/s collides with a utility pole and is brought to rest in 0.30 s. Find the magnitude of the force exerted on the car during the collision.
Stopping times and distances depend on impulse-momentum theorem Highway engineers use the theorem to
determine safe stopping distances and safe following distances
Fig. 6-3 The stopping distance and time is 2
times greater for the loaded truck than the unloaded truck.
6C Sample Problem
A 2250 kg car traveling to the west slows down uniformly from 20.0 m/s to 5.00 m/s. How long does it take the car to decelerate if the force on the car is 8450 N to the east? How far does the car travel during the deceleration?
Hint: use x = ½ (Vi + Vf)Δt from ch. 2
A change in momentum over a longer time requires less force This theorem helps design safety
equipment that exerts a less force during a collision
Ex: nets and air mattresses The change in the momentum of falling
is the same, just the net extends the time of the collision so that the change in the person’s momentum occurs over an extended time period, thus creating a smaller force on the person
Ex: fig. 6-4 and 6-5
6-2 Conservation of Momentum
Now, we’ll examine the momentum of two or more objects interacting together
Pg. 215 soccer balls
The momentum that ball B gains is exactly the same amount of momentum that ball A loses during the collision.
Table 6-1-Mass, Velocity, and Momentum
Law of Conservation of Momentum
M1v1i + m2v2i = m1v1f + m2v2f
The total momentum of all objects interacting with one another remains constant regardless of the nature of the forces between the objects.
Momentum is Conserved in Collisions
Notice that the total momentum of all objects interacting in a system are conserved but the momentum of each object is not conserved.
Frictional forces will be disregarded in most problems in this book so that most problems only have two objects interacting.
Momentum is conserved for objects pushing away from each other Ex: when you jump up, your momentum
is not conserved but when you factor in Earth, the total momentum is conserved
Fig. 6-7 Skaters Total Momentum before is zero and total momentum after is also zero
6D Sample Problem
A 76 kg boater, initially at rest in a stationary 45 kg boat, steps out of the boat and onto the dock. If the boater moves out of the boat with a velocity of 2.5 m/s to the right, what is the final velocity of the boat?
Newton’s third law leads to conservation of momentum
Remember that the force exerted by one body on another is equal in magnitude and opposite in direction to the force exerted on the first body by the second body.
F1 = -F2
The two forces act in the same time interval so that:
F1Δt = -F2Δt
The impulse on 1 is equal and opposite in magnitude to the impulse on 2
This is true in every collision between two objects
The change in momentum in the first object is equal to and opposite the change in momentum of the second object
M1v1f - m1v1i = -(m2v2f – m2v2i)
This mean if the momentum in one object increases, the momentum in the second object must decrease by that amount
Formula Rearranged
M1v1i + m2v2i = m1v1f + m2v2f
Forces in real collisions are not constant
Fig. 6-9 Although forces vary in a real collision, they are always equal but opposite in magnitude to each other
For problems, we’ll use the average force during the collision, which is equal to the constant force required to cause some change in momentum as the real, changing force.
6-3 Elastic and Inelastic Collisions
Some objects stick together and move with the momentum equal to their combined momentum before the collision
Some objects collide and bounce so that they move away with two different velocities
Total momentum is conserved in a collision but kinetic energy is generally not conserved
Some is transferred to internal energy
Perfectly Inelastic Collisions
When two objects collide and move together as one mass
The final mass is equal to the combined mass of the objects and they move with the same velocity after the collision
Formula-Inelastic Collisions
m1v1i + m2v2i = (m1 + m2)vf
Pay close attention to -/+ signs that indicate direction
+ To the right
- To the left
6E Sample Problem
A 1850 kg luxury sedan stopped at a traffic light is struck from the rear by a compact car with a mass of 975 kg. The two cars become entangled as a result of the collision. If the compact car was moving at a velocity of 22.0 m/s to the north before the collision, what is the velocity of the entangled mass after the collision?
Kinetic energy is not constant in inelastic collisions
Some of the kinetic energy is converted to sound energy and internal energy as objects deform
Elastic-means it keeps its shape Inelastic-means it is deformed and loses
some kinetic energy
6F Sample Problem
Two clay balls collide head-on in a perfectly inelastic collision. The first ball has a mass of 0.5 kg and an initial velocity of 4.00 m/s to the right. The mass of the second ball is 0.250 kg, and it has an initial velocity of 3.00 m/s to the left. What is the final velocity of the composite ball of clay after the collision? What is the decrease in kinetic energy during the collision?
Elastic Collisions
Two objects collide and return to their original shapes with no change in total kinetic energy
After the collision, the two objects move separately
In an elastic collision, the total momentum and total kinetic energy remain constant
Most collisions are neither elastic nor perfectly inelastic Most objects do not stick together and move as
one but also, most collisions result in some decrease in kinetic energy
Ex: football being kicked is deformed a little which converts kinetic energy into internal elastic potential energy
OR the formation of sound in any collision represents a decrease in kinetic energy so the collision cannot be elastic
Inelastic Collisions
Most collisions fall into this type The colliding objects bounce and move
separately after the collision, but the total kinetic energy decreases in the collision
For this book, we will assume the all collisions in which the objects do not stick will be elastic (p and KE will be constant)
KE is conserved in elastic collisions
The total momentum and total kinetic energy remain constant throughout the collision if it is a perfectly elastic collision
M1v1i + m2v2i = m1v1f + m2v2f
½ m1v1i2 + ½ m2v2i2 = ½ m1v1f2 + ½ m2v2f2
Sample Problem 6G
A 0.015 kg marble moving to the right at 0.225 m/s makes an elastic head-on collision with a 0.030 kg shooter marble moving to the left at 0.180 m/s. After the collision, the smaller marble moves to the left at 0.315 m/s. Assume that neither marble rotates before or after the collision and that both marbles are moving on a frictionless surface. What is the velocity of the 0.030 kg marble after the collision?
(remember that direction to the right is + and direction to the left is -)