# Ch7. Impulse and Momentum

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Ch7. Impulse and Momentum
There are situations in which force acting on an object is not constant, but varies with time. Two new ideas: Impulse of the force and Linear momentum of an object.

Impulse-Momentum Theorem

Definition of Impulse The impulse J of a force is the product of the average force and the time interval during which the force acts: J = Impulse is a vector quantity and has the same direction as the average force. SI Unit of Impulse: newton.second (N.s)

Definition of Linear Momentum:
The linear momentum p of an object is the product of the object’s mass m and velocity v: p=mv Linear momentum is a vector quantity that points in the same direction as the velocity. SI Unit of Linear Momentum: kilogram.meter/second(kg.m/s)

Impulse-Momentum Theorem
When a net force acts on an object, the impulse of this force is equal to the change in momentum of the object: Final momentum Initial momentum Impulse Impulse=Change in momentum

Example 1. A Well-Hit Ball
A baseball (m=0.14kg) has initial velocity of v0=-38m/s as it approaches a bat. The bat applies an average force that is much larger than the weight of the ball, and the ball departs from the bat with a final velocity of vf=+58m. Determine the impulse applied to the ball by the bat. Assuming time of contact is =1.6*10-3s, find the average force exerted on the ball by the bat.

(a) = kg.m/s (b)

Example 2. A Rain Storm Rain comes straight down with velocity of v0=-15m/s and hits the roof of a car perpendicularly. Mass of rain per second that strikes the car roof is 0.06kg/s. Assuming the rain comes to rest upon striking the car (vf=0m/s), find the average force exerted by the raindrop.

= -(0.06kg/s)(-15m/s)=0.9 N According to action-reaction law, the force exerted on the roof also has a magnitude of 0.9 N points downward: -0.9N

Conceptual Example 3. Hailstones Versus Raindrops
Suppose hail is falling. The hail comes straight down at a mass rate of m/ =0.06kg/s and an initial velocity of v0=15m/s and strikes the roof perpendicularly. Hailstones bounces off the roof. Would the force on the roof be smaller than, equal to, or greater than that in example 2? Greater.

Suppose you are standing on the edge of a dock and jump straight down. If you land on sand your stopping time is much shorter than if you land on water. Using the impulse-momentum theorem as a guide, determine which one is correct. A In bringing you to a halt, the sand exerts a greater impulse on you than does the water. B In bringing you to a halt, the sand and the water exert the same impulse on you, but the sand exerts a greater average force. C In bringing you to a halt, the sand and the water exert the same impulse on you, but the sand exerts a smaller average force. B

The Principle of Conservation of Linear Momentum

Two types of forces act on the system:
Internal forces: Forces that the objects within the system exert on each other. External forces: Forces exerted on the objects by agents external to the system. External force Internal force External force Internal force

( ) = pf - p0 m1vf1+m2vf2 = m1v01+m2v02 sum of average external forces
sum of average internal forces = pf - p0 + internal forces cancel F12 = -F21 (Sum of average external forces) = pf - p0 If sum of external forces is zero (an isolated system) Then 0 = pf - p or pf = p0 m1vf1+m2vf2 = m1v01+m2v02 pf p0

Principle of Conservation of Linear Momentum:
The total linear momentum of an isolated system remains constant(is conserved). An isolated system is one for which the vector sum of the average external forces acting on the system is zero.

Conceptual Example 4. Is the Total Momentum Conserved?
Two balls collide on the billiard table that is free of friction. Is the total momentum of the two ball system the same before and after the collision? Answer (a) for a system that contains only one ball. The total momentum is conserved. (b) The total momentum of one ball system is not conserved.

Example 5. Assembling a Freight Train
Car 1 has a mass of m1=65*103kg and moves at a velocity of v01=+0.8m/s. Car 2 has a mass of m2=92*103kg and a velocity of v02=+1.3m/s. Neglecting friction, find the common velocity vf of the cars after they become coupled.

(m1+m2) vf = m1v01 + m2v02 After collision Before collision =+1.1 m/s

Example Ice Skaters Starting from rest, two skaters push off against each other on smooth level ice (friction is negligible). One is a woman (m1=54kg), and one is a man(m2=88kg). The woman moves away with a velocity of vf1=2.5m/s. Find the recoil velocity vf2 of the man.

For the two skater system, what are the forces?
In horizontal direction F12 F21 internal forces system taken together No external forces isolated system conservation of momentum

m1vf1 + m2vf2 = 0 after pushing before pushing It is important to realize that the total linear momentum may be conserved even when the kinetic energies of the individual parts of a system change.

A canoe with two people aboard is coasting with an initial momentum of 110kg.m/s. Then person 1 dives off the back of the canoe. During this time, the net average external force acting on the system is zero. The table lists four possibilities for the final momentum of person 1 and final momentum of person 2 plus the canoe, immediately after person 1 leaves the canoe. Which possibility is correct? Person 1 Person 2 & canoe a -60kg.m/s +170kg.m/s b -30kg.m/s +110kg.m/s c -40kg.m/s -70kg.m/s d +80kg.m/s a

Collisions in One Dimension
Elastic collision: One in which the total kinetic energy of the system after the collision is equal to the total kinetic energy before the collision. Inelastic collision: One in which the total kinetic energy of the system is not the same before and after the collision; if the objects stick together after colliding, the collision is said to be completely inelastic.

Example 7. A Collision in One Dimension
A ball of mass m1=0.25kg and velocity v01=5m/s collides head-on with a ball of mass m2=0.8kg that is initially at rest(v02=0m/s). No external forces act on the balls. If the collision in elastic, what are the velocities of the balls after the collision?

(1) (2) (3) Total momentum after collision
Total momentum before collision (1) Total kinetic energy after collision Total kinetic energy before collision (2) (3)

Substitute in (2)

Substitute in (1)

m1=0.25, m2=0.8 v01 =5 m/s, v02= 0

Types of Collisions

Example 8. A Ballistic Pendulum
The ballistic pendulum consists of a block of wood(mass m2=2.5kg)suspended by a wire of negligible mass. A bullet(mass m1=0.01kg)is fired with a speed v01. After collision, the block has a speed vf and then swings to a maximum height of 0.65m above the initial position. Find the speed v01 of the bullet, assuming air resistance is negligible.

Is conservation of energy valid?
Just before collision Just after collision m1+m2 vf Is conservation of energy valid? No (completely inelastic) Is conservation of momentum valid? Yes (no external forces )

Total momentum after collision
Total momentum before collision

Applying conservation of energy
vf hf=0.65 m Applying conservation of energy Total mechanical energy at top of swing, all potential Total mechanical energy at bottom of swing, all kinetic

Two balls collide in a one-dimensional, elastic collision. The two balls constitute a system, and the net external force acting on them is zero. The table shows four possible sets of values for the initial and final momenta of the two balls as well as their initial and final kinetic energies. Which one is correct?

Initial Final a Ball 1 +4 kg.m/s 12 J -5 kg.m/s 10 J Ball 2 -3 kg.m/s
Momentum Kinetic Energy a Ball 1 +4 kg.m/s 12 J -5 kg.m/s 10 J Ball 2 -3 kg.m/s 5 J -1 kg.m/s 7 J b +7 kg.m/s 22 J +5 kg.m/s 18 J +2 kg.m/s 8 J 15 J c -6 kg.m/s -8 kg.m/s 31 J -9 kg.m/s 25 J d +9 kg.m/s +6 kg.m/s d

Example 9. A Collision in Two Dimensions

Collisions in Two Dimensions
vf2=0.7 m/s Ball 1 m1=0.15 kg Ball 2 m2=0.26 kg before after v01sin50= vf1cos 0 v02 vf2cos35 -v01cos50= vf1sin 0 -vf2sin35 x component y component

x component P0x Pfx y component P0y Pfy

Use momentum conservation to determine the magnitude and direction of the final velocity of ball 1 after the collision. x component Ball 1 after Ball 2 after Ball 2 before Ball 1 before

y component Ball 1 after Ball 2 after Ball 1 before Ball 2 before

Center of Mass

Suppose m1=5kg, m2=12kg x1=2m, x2=6m

During a time displacements of the particles, displacement of cm

m1=0.25 kg, m2=0.8 kg v01=5 m/s, v02=0 m/s Before collision After collision

Water, dripping at a constant rate from a faucet, falls to the ground. At any instant, there are many drops in the air between the faucet and the ground. Where does the center of mass of the drops lie relative to the halfway point between the faucet and the ground: above it, below it, or exactly at the halfway point? (consider gravity) Above the halfway point

Concepts & Calculations Example 10. A Scalar and a Vector

Jim and Tom are both running at a speed of 4m/s
Jim and Tom are both running at a speed of 4m/s. Jim has a mass of 90kg, and Tom has a mass of 55kg. Find the kinetic energy and momentum of the two-jogger system when Jim and Tom are both running due north. Jim is running due north and Tom is running due south.

(a)

(b)

Concepts & Calculations Example 11. Momentum and Kinetic Energy
Mass Speed Object A 2.0 kg 6.0 m/s Object B 6.0 kg 2.0 m/s Find the magnitude of the momentum and the kinetic energy for each object.

Problem 6 REASONING The impulse that the roof of the car applies to the hailstones can be found from the impulse-momentum theorem, Equation 7.4. Two forces act on the hailstones, the average force exerted by the roof, and the weight of the hailstones. Since it is assumed that is much greater than the weight of the hailstones, the net average force is equal to SOLUTION   From Equation 7.4, the impulse that the roof applies to the hailstones is:

Solving for (with up taken to be the positive direction) gives
This is the average force exerted on the hailstones by the roof of the car. The positive sign indicates that this force points upward. From Newton's third law, the average force exerted by the hailstones on the roof is equal in magnitude and opposite in direction to this force. Therefore, Force on roof = -1.8 N The negative sign indicates that this force points downward.

Problem 20 REASONING During the time that the skaters are pushing against each other, the sum of the external forces acting on the two-skater system is zero, because the weight of each skater is balanced by a corresponding normal force and friction is negligible. The skaters constitute an isolated system, so the principle of conservation of linear momentum applies. We will use this principle to find an expression for the ratio of the skater’s masses in terms of their recoil velocities. We will then obtain expressions for the recoil velocities by noting that each skater, after pushing off, comes to rest in a certain distance. The recoil velocity, acceleration, and distance are related by Equation 2.9 of the equations of kinematics.

Skater 1 glides twice as far as skater 2
Conservation of momentum Initial momentum = final momentum Ignore kinetic friction

vf1 and vf2 are just after pushing each other
vf1 and vf2 are just after pushing each other. That will be the initial velocity for the motion of the skaters and will come to rest. x1 x2 vf2 vf1 stop a1 a2 Vend=0 Vend=0 For skater 1

For skater 2 Magnitude of the acceleration is the same, i.e.,

If x1 is positive, x2 is negative,
(opposite direction) ( 1 glides twice as far as 2 )

Problem 34 REASONING AND SOLUTION Momentum is conserved in the horizontal direction during the "collision." Let the coal be object 1 and the car be object 2. Then vf The direction of the final velocity is to the right.

Problem 39 REASONING The two balls constitute the system. The tension in the wire is the only non-conservative force that acts on the ball. The tension does no work since it is perpendicular to the displacement of the ball. Since Wnc=0 J, the principle of conservation of mechanical energy holds and can be used to find the speed of the 1.50-kg ball just before the collision. Momentum is conserved during the collision, so the principle of conservation of momentum can be used to find the velocities of both balls just after the collision. Once the collision has occurred, energy conservation can be used to determine how high each ball rises.

SOLUTION a. Applying the principle of energy conservation to the 1.50-kg ball, we have m1= h0= hf=0, vf=? v0= m2= If we measure the heights from the lowest point in the swing, hf=0 m, and the expression above simplifies to Solving for vf, we have

Just before Just after v01 v02 vf1 vf2 m1 m2 Conservation of momentum

Conservation of energy

b. If we assume that the collision is elastic, then the velocities of both balls just after the collision can be obtained from Equations 7.8a and 7.8b: and Since v01 corresponds to the speed of the 1.50-kg ball just before the collision, it is equal to the quantity vf calculated in part (a). With the given values of and ,and the value of obtained in part (a), Equations 7.8a and 7.8b yield the following values: vf1= m/s The minus sign in vf1 indicates that the first ball reverses its direction as a result of the collision.

c. If we apply the conservation of mechanical energy to either ball after the collision we have
where v0 is the speed of the ball just after the collision, and hf is the final height to which the ball rises. For either ball, h0 = 0 m, and when either ball has reached its maximum height, vf = 0 m/s. Therefore, the expression of energy conservation reduces to

Thus, the heights to which each ball rises after the collision are

Problem 41 REASONING AND SOLUTION The location of the center of mass for a two-body system is given by Equation 7.10:

where the subscripts "1" and "2" refer to the earth and the moon, respectively. For convenience, we will let the center of the earth be coincident with the origin so that and , the center-to-center distance between the earth and the moon. Direct calculation then gives

Problem 50 REASONING AND SOLUTION
a. According to Equation 7.4, the impulse-momentum theorem, Since the only horizontal force exerted on the puck is the force exerted by the goalie Since the goalie catches the puck, Solving for the average force exerted on the puck, we have -2.2*103 N

By Newton’s third law, the force exerted on the goalie by the puck is equal in magnitude and opposite in direction to the force exerted on the puck by the goalie. Thus, the average force exerted on the goalie is 2.2*103N b. If, instead of catching the puck, the goalie slaps it with his stick and returns the puck straight back to the player with a velocity of –65 m/s, then the average force exerted on the puck by the goalie is -3 = -4.4*103 N

The average force exerted on the goalie by the puck is thus +4.4*103N.
The answer in part (b) is twice that in part (a). This is consistent with the conclusion of Conceptual Example 3. The change in the momentum of the puck is greater when the puck rebounds from the stick. Thus, the puck exerts a greater impulse, and hence a greater force, on the goalie.