CHAPTER 4 IMPULSE AND MOMENTUM 4.1 The Impulse – Momentum Theorem  To describe the concept of impulse, let’s refer to the situation below:  What happen when a baseball bat strikes a ball?

 

The magnitude of force is applied to the ball by the bat, where the force rises to a maximum value at the point of impact and then returns to zero when the ball leaves the bat. The time interval during which the force acts is t and the magnitude of the average force is F (refer to the figure 4.1). The product of the average force and the time of contact are called impulse.

Figure 4.1  Definition of impulse: Impulse = F t  ∆t = tf – ti  SI unit of Impulse: Ns  Impulse is a vector quantity and has the same direction as the average force.  By looking at the definition of impulse, it can be concluded that when a ball is hit, it responds to the value of impulse. A large impulse produces a large response, which the ball departs from the bat with large velocity.

 But it is also known that, if the ball is more massive, it will has less velocity after leaving the bat. So, both mass and velocity play an important role in determining the impulse. The effect between mass and velocity is introduced in the concept of linear momentum, which is defined as follows:  Definition of linear momentum: Linear momentum, p = mv  SI unit of linear momentum: kg m/s  Linear momentum is a vector quantity that points the same direction as the velocity.  Newton’s second law can be used to reveal a relationship between impulse and momentum. How??  Refer figure 4.2: When a bat hits a ball, average force is applied to the ball by the bat. As a result, the ball’s velocity changes from initial value of vo to a final value of vf.

Figure 4.2



When the velocity of the ball changes from vo to vf, the average acceleration is: v f  vo a t



But according to Newton’s second law, F = ma. Thus  v f  vo  mv f  mvo  F  m  t  t  Multiplying both side of the equation above by t yields a theorem known as impulse – momentum theorem.  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. (F) t = mvf Impulse

mvo

Initial Final momentum momentum

Impulse = change in momentum

EXAMPLE 4.1 A golf ball of mass 0.045kg is hit off the tee at a speed of 45m/s. The golf club was in contact with the ball for 3.5 × 10-3 s. Find a) the impulse imparted to the golf ball, and b) the average force exerted on the ball by the golf club. Solutions: (a) The impulse is the change in momentum. The direction of travel of the struck ball is the positive direction.





p  mv  4.5 102 kg  45 m s  0   2.0 kg m s

(b)

The average force is the impulse divided by the interaction time. p 2.0 kg m s F   5.8  102 N t 3.5 103 s

EXAMPLE 4.2 A 12kg hammer strikes a nail at a velocity of 8.5 m/s and comes to rest in a time interval of 8.0 ms. a) what is the impulse given to the nail? b) What is the average force acting on the nail? Solutions (a) The impulse given to the nail is the opposite of the impulse given to the hammer. This is the change in momentum. Call the direction of the initial velocity of the hammer the positive direction. pnail  phammer  mvi  mv f  12 kg  8.5 m s   0  1.0 102 kg m s

(b)

The average force is the impulse divided by the time of contact. p 1.0  102 kg m s Favg    1.3  104 N t 8.0  103 s

4.2 The Principle of Conservation of Linear Momentum  The impulse – momentum theorem leads to the principle of conservation of linear momentum, when the sum of average external force acting on the system is zero.  Recap the impulse – momentum theorem:  (Sum of average external forces) t = pfinal - pinitial  If the sum of the average external forces is zero, then pfinal = pinitial which is the mathematical statement of conservation of linear momentum.  Example of conservation of linear momentum: Refer to figure 7.3, the total linear momentum of the balls before the collision is equal to the total linear momentum after the collision.

Figure

 Principle of conservation of Linear Momentum: The total linear momentum of an isolated system remains constant (is conserved), which the vector sum of the average external forces acting on the system is zero. EXAMPLE 4.3 A 9300kg boxcar traveling at 15.0m/s strikes a second boxcar at rest. The two stick together and move off with a speed of 6.0m/s. What is the mass of the second car? Solution: Consider the motion in one dimension, with the positive direction being the direction of motion of the first car. Let “A” represent the first car, and “B” represent the second car. Momentum will be conserved in the collision. Note that vB  0 . pinitial  pfinal  mAvA  mB vB   mA  mB  v 

mB 

mA  vA  v   9300 kg 15.0 m s  6.0 m s    1.4  104 kg v 6.0 m s

4.3 Collisions in One Dimension  There are two types of collisions:  Elastic collision – Total kinetic energy of the system after the collision is equal to the total kinetic energy before the collision. 

Inelastic collision – Total kinetic energy of the system is not same before and after the collision. But if the objects stick together after colliding and move off with same velocity, the collision is said as perfectly / completely inelastic. Greatest amount of KE is lost when the collision is completely inelastic.

 Refer to figure 4.4 to get a clearly view regarding types of collisions that are mentioned above.

 4.4(a): If the collision is elastic, the steel ball will rebound to its original height.

4.4

4.4

 4.4(b): If the collision is inelastic, a partially deflated ball has a little bounce on the surface,

indicating that the KE is dissipated during the collision.

 4.4(c): If the collision is completely inelastic, the deflated ball has no bounce at all, and a maximum of KE is lost during the collision. 4 .

Elastic collisions: EXAMPLE 4.4 A ball of mass 0.440kg moving east (+x direction) with a speed of 3.30 m/s collides head-on with a 0.220kg ball at rest. If the collision is perfectly elastic, what will be the speed and direction of each ball after the collisions? Solutions: Let A represent the 0.440-kg ball, and B represent the 0.220-kg ball. We have vA  3.30 m s and

vB  0 . Relationship between the velocities:

vA  vB    vA  vB   vB  vA  vA Substitute this relationship into the momentum conservation equation for the collision. mA vA  mB vB  mA vA  mBvB  mA vA  mA vA  mB  vA  vA  

vA 

0.220 kg  mA  mB  vA   3.30 m s   1.10 m s  east  0.660 kg  mA  mB 

vB  vA  vA  3.30 m s  1.10 m s  4.40 m s  east 

EXAMPLE 4.5 Two billiard balls of equal mass undergo a perfectly elastic head-on collision. If one ball’s initial speed was 2.00 m/s, and the other’s was 3.00 m/s in the opposite direction, what will be their speeds after the collision? Solutions:` Let A represent the ball moving at 2.00 m/s, and call that direction the positive direction. Let B represent the ball moving at 3.00 m/s in the opposite direction. So vA  2.00 m s and

vB  3.00 m s . Relationship between the velocities. vA  vB    vA  vB   vB  5.00 m s  vA Substitute this relationship into the momentum conservation equation for the collision, noting that mA  mB .

mA vA  mB vB  mA vA  mBvB  vA  vB  vA  vB  1.00 m s  vA   vA  5.00 m s   2vA  6.00 m s  vA  3.00 m s vB  5.00 m s  vA  2.00 m s

The two balls have exchanged velocities. This will always be true for 1-D elastic collisions of objects of equal mass.

4.4 Collisions in Two Dimensions  Example of collisions in two dimensions: Refer figure 4.5, two balls colliding on a

vfiy vfix

horizontal frictionless table.  In 2-D, the x and y components of the total momentum is conserved separately, or in other words, the equations can be described as: x  component : m1v f 1x  m2v f 2 x  m1vo1x  m2vo 2 x Pfx

Pox

y  component : m1v f 1 y  m2v f 2 y  m1v01 y  m2vo 2 y Pfy

Poy

 Remember that momentum is a vector quantity, so we have to take into account the direction as well. The direction can be determined by using the vector component, shown in figure 4.5(b). EXAMPLE 4.7

An eagle (mA = 4.3kg) moving with a speed vA = 7.8m/s is on a collision course with a second eagle (mB = 5.6kg) moving at vB = 10.2 m/s in a direction perpendicular to the first. After they collide, they hold onto one another. In what direction, and with what speed, are they moving after the collision? Solution: Consider the diagram for the momenta of the eagles. Momentum will be conserved in both the x and y directions.

py   mA  mB  vy  mBvB  vy 

2

mA  mB mA  mB

 4.3 kg   7.8 m s    5.6 kg  10.2 m s 



2

4.3 kg  5.6 kg

2



pB  mB v B

 mA vA    mBvB 

2

2

2 y

2

pA  mA v A

mBvB

 mv   mv  v  vx  v   A A    B B    mA  mB   mA  mB  2

p   mA  mB  v

mA vA

px   mA  mB  vx  mA vA  vx 

mA  mB 2

 6.7 m s

2