Chapter 7 - Impulse and Momentum PDF

Preview

Summary

Chapter summary and practice...

Description

CHAPTER 7

IMPULSE and MOMENTUM

7.1 The Impulse-Momentum Theorem From our study of Newton’s Laws of Motion, a force must be present to change an object’s velocity whether its speed or direction that is changed. To more fully understand the impact of a force acting on an object, it is useful to introduce the concept of momentum. The linear momentum of an object is defined as the product of the object’s mass   times its velocity: p  mv Linear momentum is a vector quantity, is in the same direction as the velocity, and has units of kilogram ∙ meter/second (kg ∙ m/s). Note that linear momentum of an object can change if either its mass or velocity changes.

When the bat strikes the baseball, there is a force exerted on the baseball over a time t, as shown below. Note that the force is not necessarily constant over the time of the impact.

The impulse of this force is defined as

  impulse  J   F  dt  area under F-vs-t curve   impulse  J  Favet

Impulse is a vector quantity, is in the same direction as the force, and has units of Newton ∙ seconds (N ∙ s).

For given impulse, one would expect that change in velocities of the object before and after the collision will depend upon the mass of the object. For example, compare a baseball and a bowling ball with the same initial speeds.

Thus when a net force acts on an object, the impulse of this force is equal to the change in the linear momentum of the object

     impulse  Fave t  mv f  mvo  pf  po

Derivation:

final momentum initial momentum      vf  vo Fave   F  ma  m t      Fave t  mv f  mv o  p f  p o    Fave t  p f  po

Example 1 If a 0.145-kg baseball traveling at 36 m/s is hit by 0.950-kg baseball bat, find the impulse imparted to the baseball if the baseball leaves the bat with a velocity of 40 m/s.

Assuming that bat is in contact with the baseball for 0.7 ms, find the average force acting on the baseball.

According to Newton’s 3rd Law, what is the reaction force to the force felt by the baseball? the reaction force is the force of the baseball on the bat FR   F  t   FR  t     mbaseball(v f  v o )   mbat (v 'f  v 'o )     m baseballv f  m batv'f  m baseballv o  m batv'o     pf (baseball)  pf (bat)  po (baseball)  po ( bat) Conservation of linear momentum 7.2 The Principle of Conservation of Linear Momentum The principle of conservation of linear momentum states that total linear momentum of an isolated system is a constant (conserved) when no net external force acts on the system of two or more objects. That is, the total momentum of the system before the collision is equal to the total momentum of the system after the collision.

   ( Fave )external t  pf (total )  po (total )  if ( Fave ) external  0   then p o( total)  p f ( total)

Internal forces – Forces that objects within the system exert on each other. External forces – Forces exerted on objects by agents external to the system. Note another object would have to influence the collision by creating an external force acting on the objects involved in the collision.  

po ( tota l)  pf ( tota l)     m1v o1  m2v o2    m1v f1  m2v f 2 

Returning to Example 1 with the baseball and bat, find the change in velocity of bat.

Example 2 Ice Skaters Starting from rest, two skaters push off against each other on ice where friction is negligible. One is a 45-kg woman and one is a 90-kg man. The woman moves away with a speed of +2.5 m/s. Find the recoil velocity of the man.

Applying the Principle of Conservation of Linear Momentum 1. Decide which objects are included in the system. 2. Relative to the system, identify the internal and external forces. 3. Verify that the system is isolated. 4. Set the final momentum of the system equal to its initial momentum. Remember that momentum is a vector. 7.3 Collisions in One Dimension Total linear momentum is conserved when two objects collide, provided they constitute an isolated system. 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, i.e., total KE before = total KE after Inelastic collision -- One in which the total kinetic energy of the system after the collision is not equal to the total kinetic energy before the collision; if the objects stick together after colliding, the collision is said to be completely inelastic. total KE before = total KE after + energy lost

Example 3 A 1000-kg car auto traveling at 50 m/s crashes into a parked 1500-kg car. After the collision, the twisted wreckage remains joined together. Find the velocity of wrecked cars after the collision. How much kinetic energy is “lost” (i.e., converted to heat, the twisting of metal, etc.) in the wreck?

Example 4 A 1000-kg car auto traveling at 50 m/s crashes into the1500-kg car which is traveling 20 m/s in the opposite direction. After the collision, the 1000-kg car is traveling at a speed of +5 m/s (to the right). Find the velocity of the1500-kg car after the collision.

Example 5 A 1000-kg car auto traveling at 50 m/s crashes into the1500-kg car which is traveling 20 m/s in the opposite direction. Find the velocities of the two cars after a perfectly elastic collision.

Example 6 If the mass of both objects is the same, the velocities of the two objects are interchanged after a perfectly elastic collision.

Example 7 A Ballistic Pendulum The mass of the block of wood is 2.50-kg and the mass of the bullet is 0.0100-kg. The block swings to a maximum height of 0.650 m above the initial position. Find the initial speed of the bullet.

7.4 Collisions in Two Dimensions Most collisions are not head-on or 1-dimensional, but two-dimensional with the objects moving at angles.

Linear momentum is still conserved.   p befo re( total)  p after ( total)     m1 v o1  m2 v o2  m1 v f1  m2 v f2 where momentum is conserved in both the x & y directions m1v o1 x  m2 vo2 x  m1v f 1 x  m2v f 2 x m1 vo1 y  m2 vo2 y  m1 v f 1 y  m2 v f 2 y

Example 8 For the collision shown at the right for equal masses, which of the following scenarios after the collision could occur based on momentum conservation?

Example 9 An open box slides across a frictionless surface. What happens to the speed of the box as water from a rain shower collects in the box, assuming the rain falls vertically downward into the box?

Example 10 A boy stands at one end of a floating raft that is stationary relative to the shore. He then walks to the opposite end, towards the shore. Does the raft move? Assume no friction between the water and the raft.

7.5 Center of Mass Up to now, we have ignored the size of an object when analyzing the motion of a large object as we treated its motion as if it was a point particle. Although large bodies consist of a large number of particles with each particle obeying the law: Δp/Δt = Fext describing their motion becomes very complicated because we have to find the net momentum for all of the particles in the body. One can greatly simplify things by reducing the motion of this system of particles to the motion of one single point for the object: its Center of Mass The position of the Center of Mass is merely the average position of the mass of the system, i.e., it is the average position of all of the small volume elements making up the volume of the body. The center of mass is a point that represents the average location for the total mass of a system.

m1x1  m 2 x 2 m1  m 2 m y  m2 y2 ycm  1 1 m1  m 2 m z  m2 z 2 z cm  1 1 m1  m 2

xcm 

As the center of mass changes its location over a time interval t, one can describe the velocity of the center of mass vcm.

xcm 

vcm 

m1 x1  m2 x2 m1  m2

m1v1  m2 v2 m1  m2

Since the total linear momentum does not change for an isolated system (no net external force), the velocity of the center of mass does not change.

Example 2 Ice Skaters Starting from rest, two skaters push off against each other on ice where friction is negligible. One is a 45-kg woman and one is a 90-kg man. The woman moves away with a speed of +2.5 m/s while the recoil velocity of the man was determine to be 1.25 m/s. Determine the velocity of the center of mass before and after. BEFORE

vcm 

m1 v1  m2 v2 0 m1  m 2

AFTER

vcm  vcm

90 kg 1.25 m s   45 kg  2.5 m s 

90 kg  45 kg 112.5 kg m s   112.5 kg  m s   0  135 kg...