Week 12 ---Impulse+Momentum+Impact

ImpulseMomentum

Impact

Impulse & Momentum

Analysis velocity of objects subjected to external forces and impacts.

Principle of Linear Impulse & Momentum

Consider the curvilinear motion of a particle of mass m.

The velocity is in a tangential direction to the path.

rv

The resultant force will be in the direction of acceleration.

dtdvmvmma F

Principle of Impulse & Momentum

dtdG

dtmvd

dtdvm )(F

mvG

GF The resultant force acting on a particle equals its rate of change of linear momentum.

The product of the mass m and the velocity v is defined as the linear momentum.

LINEAR MOMENTUM AND IMPULSE

Linear impulse: The integral F dt is the linear impulse, denoted I. It is a vector quantity measuring the effect of a force during its time interval of action. I acts in the same direction as F and has units of N·s.

Linear momentum: The vector mv is called the linear momentum, denoted as G. This vector has the same direction as v. The linear momentum vector has units of (kg·m)/s.

Principle of Impulse & Momentum

dtdG

dtmvd

dtdvm )(F G Fdt

12

2

1

GGFdtt

t

12

2

1

mvmvFdtt

t

21

2

1

GFdtGt

t

Initial linear momentum + Linear impulse

= Final momentum

Principle of Impulse & Momentum

xxxx

t

tx mvmvGGdtF 1212

2

1

The three components in x, y and z are

yyyy

t

ty mvmvGGdtF 1212

2

1

zzzz

t

tz mvmvGGdtF 1212

2

1

Principle of Impulse & Momentum

x

t

txx mvdtFmv 21

2

1

Initial linear momentum + Linear impulse

= Final momentum

y

t

tyy mvdtFmv 21

2

1

z

t

tzz mvdtFmv 21

2

1

Principle of Impulse & Momentum

• It states that the impulse applied to an object during an interval of time is equal to the change in the object’s linear momentum.

12

2

1

mvmvFdtt

t

Procedure for Analysis

• Establish the x, y, z coordinate system.

• Draw the particle’s free body diagram and establish the direction of the particle’s initial and final velocities, drawing the impulse and momentum diagrams for the particle. Show the linear momentum and force impulse vectors.

Procedure for Analysis

• Forces as functions of time must be integrated to obtain impulses. If a force is constant, its impulse is the product of the force’s magnitude and time interval over which it acts.

• Resolve the force and velocity (or impulse and momentum) vectors into their x, y, z components, and apply the principle of linear impulse and momentum using its scalar form.

EXAMPLE 1The 100-kg stone is originally at rest on the smoothhorizontally surface. If a towing force of 200 N,acting at an angle of 45°, is applied to the stone for10 s, determine the final velocity and the normalforce which the surface exerts on the stone duringthe time interval.

EXAMPLE 1Free Body Diagram

Impulse )( 12 ttFI c

Velocity: horizontal and positive when pointing to the right

EXAMPLE 1

EXAMPLE 1

EXAMPLE 2The 250-N crate is acted upon by a force having avariable magnitude P = (100t) N. Determine thecrate’s velocity 2 s after P has been applied. Theinitial velocity is v1 = 1 m/s down the plane, and thecoefficient of kinetic friction between the crate andthe plane is μk = 0.3.

EXAMPLE 2Free Body Diagram

Impulse

--- For varying force P:integrating P = 100t overthe 2-s time interval.

---For constant forces(Weight, normal forceand frictional force): )( 12 ttFI c

EXAMPLE 2

EXAMPLE 2

EXAMPLE 3Given: A 40 g golf ball is hit over

a time interval of 3 ms by a driver. The ball leaves with a velocity of 35 m/s, at an angle of 40°. Neglect the ball’s weight while it is struck.

Find: The average impulsive force exerted on the ball and the momentum of the ball 1 s after it leaves the club face.

EXAMPLE 3Plan: 1) Draw the momentum and impulsive

diagrams of the ball as it is struck.2) Apply the principle of impulse and

momentum to determine the average impulsive force.

3) Use kinematic relations to determine the velocity of the ball after 1 s. Then calculate the linear momentum.

Solution:1) The impulse and momentum diagrams can be

drawn:

The impulse caused by the ball’s weight and the normal force N can be neglected because their magnitudes are very small as compared to the impulse of the club. Since the initial velocity (vO) is zero, the impulse from the driver must be in the direction of the final velocity (v1).

mvO = 0

+ =

W dt 0

N dt 0 F dt

mv1

40°

2) The principle of impulse and momentum can be applied along the direction of motion:

3) After impact, the ball acts as a projectile undergoing free-flight motion. Using the constant acceleration equations for projectile motion:

Considering two particles interacting during a time interval t, if the interaction forces are F one particle.

Conservation of Linear Momentum

Then it will be F for the other particle.

FtG 1

FtG 2

Conservation of Linear Momentum

Therefore the total change in linear momentum for the system is

0)( FtFt

21 GG

0G

or

This is known as the principle of conservation of linear momentum.

For the system of particles shown, the internal forces fibetween particles always occur in pairs with equal magnitude and opposite directions. Thus the internal impulses sum to zero.

Principle of Linear Impulse & Momentum for a System of Particles

The internal forces acting between particles do not appear with this summation, since by Newton’s third law they occur in equal but opposite collinear pairs and therefore cancel out.

Principle of Linear Impulse & Momentum for a System of Particles

dtdvmvmamF i

iiiiii

The linear impulse and momentum equation for this system only includes the impulse of external forces.

Principle of Linear Impulse & Momentum for a System of Particles

21

2

1

ii

t

tiii vmdtFvm

EXAMPLE 4Given:Two rail cars with masses of mA = 15 Mg and

mB = 12 Mg and velocities as shown.

Find: The speed of the cars after they meet and connect. Also find the average impulsive force between the cars if the coupling takes place in 0.8 s.

Plan: Use conservation of linear momentum to find the velocity of the two cars after connection (all internal impulses cancel). Then use the principle of impulse and momentum to find the impulsive force by looking at only one car.

Solution:

Impact

Impact occurs whentwo bodies collidewith each otherduring a very shortperiod of time,causing relativelylarge (impulsive)forces to be exertedbetween the bodies.

ImpactThe line of impact passes through the mass centers of the particles.Central impact: thedirection of motion of themass centers of the twocolliding particles isalong the line of impact.

Oblique impact: one orboth of the particles ismoving at an angle withthe line of impact.

Central Impact• Consider two smooth particles A and B with, theinitial momentum as shown• Provided , collision will Eventually occur.

11 )()( BA vv

The particles are deformable or non-rigid duringthe collision.

• An equal but opposite deformation impulse ∫ Pdtis exerted on each other.

Central Impact

At the instant of maximum deformation• Their relative motion is zero• Both particles move with a common velocity v.

Central Impact

• Afterward a period of restitution occurs (theparticles will either return to their original shapeor remain permanently deformed).• The equal but opposite restitution impulse ∫R dtpushes the particle apart from one another.

Central Impact

Just after the separation the particles will havethe final momentum, where 22 )()( AB vv

Central Impact

Conservation of linear momentum

Central Impact

2)( AvWith two unknown velocities 2)( Bvand

2211 )()()()( BBAABBAA vmvmvmvm

The coefficient of restitution.

Central Impact

11

22

)()()()(

BA

AB

vvvve

22 )()( AB vv

Relative velocity just after impact

11 )()( BA vv

Relative velocity just before impact

Coefficient of restitution e has a value between 0and 1 depending on the material property of the particles

Central Impact

Find the final velocities just after central impact.

Procedure for Analysis-Central Impact

Given:• Coefficient of restitution,• Mass of each particle• Initial velocity of each particle just before impact

Equations:• The conservation of momentum: Σmv1 = Σmv2• coefficient of restitution relating the relativevelocities before and just after collision.

EXAMPLE 5The bag A, having a mass of 6 kg is released fromrest at the position θ = 0°. After falling to θ = 90°, isstrikes an 18 kg box B. If the coefficient ofrestitution between the bag and the box is e = 0.5,determine the velocities of the bag and box justafter impact and the loss of energy during collision.

Solution

Coefficient of Restitution

Oblique ImpactParticles travel at an angle with the line of impact.

The equations for centric impact are still valid for oblique impact. We need to work out

2)( Av 2)( Bv 2 2 or 2)( Axv 2)( Ayv 2)( Bxv 2)( Byv

Choose the line of impact as the x axis.

Procedure for Analysis-Oblique Impact

The impulsive forces of deformation and restitution act only in the x direction.

2)( Axv 2)( Ayv 2)( Bxv 2)( Byv

Resolving the velocities into components along x and yaxes.

Conservation of momentum along x axis.

Procedure for Analysis-Oblique Impact

Coefficient of restitution

21 )()( xx vv

11

22

)()()()(

BxAx

AxBx

vvvve

Conservation of momentum along y axis (no impulse on A and B)

Procedure for Analysis-Oblique Impact

21 )()( ByBByB vmvm

21 )()( AyAAyA vmvm

EXAMPLE 6Two smooth disks A and B, having mass of 1 kgand 2 kg respectively, collide with the velocitiesshown. If the coefficient of restitution for the disksis e = 0.75, determine the x and y components ofthe final velocity of each disk just after collision.

Solution

Resolving each of the initial velocities into x and ycomponents, we have

Conservation of “x” Momentum

The four unknown velocity components after collision are assumed to act in the positive directions.

Coefficient of (x) Restitution. Both disks are assumed to have components of velocity in the +x direction after collision

Conservation of “y” Momentum