Chapter Part 1 Momentum, Impulse, Collisions

Prof. Clements Notes/Hints for Physics Courses

Chapter 8 Part 1 Open Stax College Physics or Most Introductory Physics Courses

Momentum, Impulse, Collisions

Terms you should know at the end of this unit: momentum, internal force, external force, impulse, conservation of momentum, collision, KE, elastic collision, inelastic collision

Linear Momentum and Force

Your studies in this chapter will inform you about the second great conservation law of physics, conservation of momentum. Momentum is a very useful quantity for the analysis of motion of objects. It is calculated by multiplying the mass and velocity for an object. The symbol for momentum is p.

p = mv The units of momentum are kg m/sec. Momentum is a vector that has the same direction as the velocity of the object. The form of Newton’s Second Law that we have been using is actually a special case of a more general law. F_net = Δ p/ Δ t

The text shows how F = ma is derived from the general law.

F = ma is not the form of the original “Newton’s Second Law.”

F = change in momentum / time interval is the original form of Newton’s second law.

When a rocket is launched the equation F = ma is not directly useful for finding the acceleration so we cannot just use the four kinematic equations to determine the rocket’s motion. The mass will be changing as fuel is consumed so the acceleration is not constant. Also, the rate of use of fuel (and thus the force) is variable for real rocket launches.

In a system that consists of more than one object, the total momentum is found by adding the separate momenta of each object.

Impulse

Impulse is found by multiplying the average force by the length of time that the force was acting. Impulse is equal to the change in momentum. F Δt = Δp

The Force will be the average, or effective, force during the time interval.

e.g. Two objects are thrown with a speed of 5 m/s toward a wall. Ball A stops when it hits the wall and does not rebound. Ball B rebounds. Does the wall exert the same force on the two balls?

Answer The impulse on Ball A is greater than the impulse on Ball B. You may assume the contact times are equal for the two collisions. The momentum for Ball A went from some value to zero. The change in momentum for ball B is twice as large (due to the rebound) as the change in momentum for ball A so the force on B is larger by a factor of 2. Also, consider the work done by the forces. The force on ball A has just stopped the ball. The force on ball B has stopped the ball and then accelerated it away from the wall.

The use of “shock absorbers” in car bumpers is a safety and car care device. In a low speed (e.g. 5 mile/hour or about 10 m/s) accident the shock absorber creates a longer time of collision which leads to a smaller contact force and less damage to the bumper.

Think about the equation F_net = Δ p/ Δ t and try to write an explanation of the importance of bending your knees when you land after jumping.

Describe the hitting technique that causes the greatest change in momentum for a tennis ball.

Answer: Follow through the swing. This will lead to longer contact time and more impulse.

Conservation of Momentum

If the external forces add to zero then the total momentum of the system will be conserved (p_total will be a constant). Consider F_net = Δ p/ Δ t . If the net force is zero then Δ p is zero. The momentum has not changed. We will use the expression p_final - p_initial = 0 or, p_final = p_initial

The total momentum of an isolated system of bodies remains constant.

Q6. A halfback who weighs 180 pounds launches himself towards the goal line with the football at 10 m/sec. A 220 pound linebacker launches himself towards the halfback at 7 m/sec. They collide such that the football is just a fraction of an inch away from the goal line.

Will the ball cross the goal line?

Answer: Each player has momentum. When they meet in the air, instead of having their feet pushing on the ground, the external force is zero. If one or both players push on the ground you cannot use conservation of momentum to solve this problem. One of the players has positive momentum and the other negative as one of the velocities will be in the negative direction. You don’t need to convert units. Mass is proportional to weight on the Earth. You must use the same units on both sides of the conservation of momentum equation.

180 lb *10 m/s - 220 lb * 7 m/s = (180 lb + 220 lb) * V_final

V_final = + 0.65 m/s . The velocity of the two person lump is positive so the touchdown is scored.

Questions for thought:

Suppose that a car crash on a street occurs and both drivers apply their brakes during the collision. Is momentum conserved for this collision? Explain.

Answer: No. The net external force is not zero due to friction of the road on the cars.

Why does a rifle recoil when it is shot?

Answer: The momentum of the system before the rifle is shot is zero. The explosion in the chamber that pushes the bullet out the rifle is an internal force. Assume that the rifle is only lightly held so the external force is about zero. The total momentum of the bullet and the rifle must be zero after the shot is fired. The bullet goes in one direction with positive momentum. The rifle moves in the other direction with negative momentum.

Elastic Collisions in One Dimension

There are three types of collisions that occur between objects:

1) elastic (also called perfectly elastic) KE of the system is conserved eg. atomic collisions;

2) inelastic some KE is lost e.g. kicking a football, hitting a tennis ball;

3) completely inelastic much KE is lost e.g. a lineman tackles a halfback in football; two train cars link together; air track cars with velcro strips stick together; a ball is caught by a pendulum.

If the external forces can be ignored, momentum is conserved in all types of collisions. This fact makes the momentum quantity very useful in analyzing collisions.

In a collision between a car and a truck, the force of the car on the truck is the same value as the force of the truck on the car. The forces have the same magnitude. This is a Newton’s Third Law concept.

When the equations of conservation of momentum and conservation of kinetic energy are combined we find that

V_1new = V₁ * (m₁ – m₂) / (m₁ + m₂) + 2 * V₂ * m₂ / (m₁ + m₂)

V_2new = 2*V₁*m₁ / (m₁ + m₂) - V₂ * (m₁ – m₂) / (m₁ + m₂)

You should deduce the answers to the following. Ask your instructor if you have questions.

How do these equations simplify for the case of V₂ = 0 (target at rest) and

a) m₁ = m₂ ? b) m₁ >> m₂ ? c) m₁ << m₂ ?

What will happen if a pair of elastic balls are dropped in line onto a hard surface? The balls are dropped at the same time and the top object collides directly with the top of the bottom object as the bottom object rebounds from the hard surface.  Let the mass of the top ball be very small compared to the mass of the bottom ball.  If the balls have a speed V when approaching the hard surface, what is the speed of the top ball as it leaves the collision?

Try this demonstration on your own and share your observations with someone in your physics class: Lay a marble in the groove of a plastic ruler (or just on the table).

a) Shoot a marble with similar mass at the stationary marble. Observe the collision.

b) Shoot a massive steel ball at a stationary marble. Observe the collision.

c) Shoot a marble at stationary, massive, steel ball. Observe the collision.

Read about the operation of “Newton’s Cradle” in your textbook or on the Internet. This item is often seen in science gift shops at museums.

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