12. Conservation of MomentumRevFF.pdf

 At the end of this chapter, you should be able to:

1. Define Linear Momentum;

2. State the law of conservation of linear momentum

and apply it to solve some dynamics problems;

3. Define and classify Collisions;

4. Apply the law of conservation of linear

momentum to solve problems involving collisions; and

5. Define Impulse and the Impulse-Momentum

 Linear Momentum

 _{1. Linear Momentum}

 _{2. Momentum and Newton's Second Law}  _{3. Conservation of Momentum}

 Collisions

 _{4. Overview ~ Collisions}

 _{5. Elastic and Inelastic Collisions}  _{6. Impulse of a Force}

Can be thought of as the effort you need to stop an object from moving.

Determined by two factors:

1. The object’s inertia (mass)

2. The object’s velocity

For example, a heavy truck has more momentum than a light car travelling at the same speed.

 A particle's momentum p is defined as the

product of its mass and velocity

p

= m

v

 Physical Properties  Symbol: p

 Newton's Second Law can be written in terms of

the momentum of a particle.

 Recall Newton’s Second Law:

 Therefore we can write the 2nd law as:

 _{Recall that p = mv, so Δp = mΔv}

 Thus the net force acting on a particle equals the

Choose the direction from the batter to the pitcher to be the + direction. Given: m1 = 0.145 Kg; V1 = 39 m/s; V2 = 52 m/s; ∆t = 3 x10-³s

Find : average force from the change in momentum of the ball.

Solution:

∆p = F ∆t = m ∆V; F = m ∆V/ ∆t = m(V2 –V1)/∆t = ((0.145Kg)(52m/s – (-39m/s))/3 x10-³s

 The total momentum P of a system of particles

is the sum of the momenta of the individual particles.

p = Σ m_iv_i = Σ p_i

 According to Newton's Second Law,

 Requirement: When the net external force

acting on a system of particles is zero,

 The rate of change of total momentum is zero;

and the total momentum of the system remains constant:

P = Σ m_iv_i = constant (F_net,ext = 0)

 This is known as the law of conservation of

 Like energy, momentum is state-based.

 With this, we can express the conservation of

momentum as



P

= P

4. Given: mass of child = 26Kg; mass of the boat = 45 Kg; mass of package thrown = 6.4 Kg at a speed of 10m/s

Find: speed of both child and boat after throwing the package

Solution:

The throwing of the package is a momentum-conserving action, if the water resistance is ignored.

To simplify the analysis, Let “a” represent the boat and child together, and let “b” represent the package. Choose the direction that the package is thrown as the positive direction.

Pi = Pf ; (Ma + Mb)V = MaVa’ + Mb Vb’

With the initial velocity of both objects being 0; .V = 0 so MaVa’ = - MbVb’; Va’ = - MbVb’/Ma

 _{Va’ = - (6.40 kg )(10.0m/s)/(26Kg + 45Kg) = - 0.901 m/s}

12. Given: mass of bullet = 23 g; Velocity of bullet = 230 m/s; mass of block = 2 Kg and Velocity of bullet after penetration = 170 m/s

Find: Velocity of block

Solution

Consider the motion in one dimension with the positive direction being the direction of motion of the bullet. Let “a” represent the bullet, and “b” represent the block. Since there is no net force outside of the block-bullet system (like frictions with the table), the momentum of the block and

bullet combination is conserved.

Pi = Pf = Ma Va + Mb Vb = MaVa’ + MbVb’ Note that Vb = 0 as the block is at rest

So Vb’ = Ma( Va - Va’)/Mb = 0.023 Kg (230 m/s – 170 m/s) / 2 Kg

 In a collision, two objects approach and interact strongly for a very short time.

 During this brief time of collision,

F _ext << F _{interaction between two objects}

 So the total momentum of the system remains

unchanged. (Recall 3rd Law)

 The collision time is usually so small that the

 Before and after the

collision, the interaction of the two objects is

 When the total kinetic energy of the objects is the same after collision as before the collision is called an elastic collision

 _{The effect of this is the velocities of the two}

 When the total kinetic energy of the objects is not the same, it is termed an inelastic collision.

 _{An extreme case is the perfectly inelastic collision,}

 _{Both momentum and kinetic energy are conserved}  Typically have two unknowns

 _{Solve the equations simultaneously}

 _{Draw “before” and} “after” sketches

 _{Label each object}

 include the direction of velocity

Mass 1 = 1000Kg; Mass 2 = 1500Kg V1 = 50 m/s; V2 = - 20 m/s

s m kg s m kg s m kg v m v m

Pbefore i i

        4 2 2 1 1 10 0 . 2 ) 20 )( 1500 ( ) 50 )( 1000 ( s m v s m v f f 1 . 31 7 . 26 2 1    

For perfectly elastic collision:

J J J v m v m

KE_{before i i}

6 5 6 2 2 2 2 1 1 10 55 . 1 10 3 10 25 . 1 2 1 2 1         2 2 2 2 1 1 6 2 2 1 1 4 2 1 2 1 10 55 . 1 10 0 . 2 f f f f v m v m J v m v m s m kg         S O L V E T H I S

 _{When two objects} stick together after the collision, they have undergone a

perfectly inelastic collision

f i

i m v m m v

v

m_{1 1}  _{2 2}  ( ₁  ₂)

f

i m m v

v

 Momentum = mass x velocity

 Momentum is conserved for a system when the

net external force on that system is 0.

 Collisions

 Impulse is associated with the forces of

interaction during collisions.

 During the time of collision(very small), the

Physical Properties:

 Symbol: _I

 Type: Derived, Vector Quantity  Formula:

I = F Δt = F (t_f – t_i)

 Dimension [F*T]; SI Units: N*s

(Newton*second)

I

= F

Δt

I

=

Δp = p

– p

 The average force for the time interval t_f – t_i is defined

as

 _{The average force is the constant force that gives}

the same impulse as the actual force in the time interval Δt.

 _{This time is often estimated using the distance}

35

36 In a head-on collision:

Newton's third law dictates that the forces on the trucks are equal but opposite in direction.

Impulse is force multiplied by time, and time of

contact is the same for both, so the impulse is the same in magnitude for the two trucks. Change in momentum is equal to impulse, so changes in momenta are equal.

 16. (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.

a) ∆p (nail) = -∆p (hammer) = mvi - mvf = 12Kg (8.5m/s) – 0 = 102 Kg m/s

(b) The average force is the impulse divided by the time of contact.

a)Consider an elastic collision between masses

m1 = 100g and m2 = 200g. Ball 1 is moving with a velocity of 30cm/s to the right and ball 2 has a

velocity of 20 cm/s, also to the right. Find the final velocity of the two balls after collision

b) A car (mass, 1200 kg) with an initial constant velocity of 35 km/h collides head-on with a truck (mass, 14,000 kg), with a velocity of 13 km/h. If