Impulse Momentum Impact
Impulse & Momentum Analysis velocity of objects subjected to external forces and impacts.
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.
is in a The velocity v r tangential direction to the path. The resultant force will be in the direction of acceleration.
dv
F ma mv m dt
Principle of Impulse & Momentum dv
F m dt
d ( mv ) dt
dG dt
The product of the mass m and the velocity v is defined as the linear momentum.
G mv
F G
The resultant force acting on a particle equals its rate of change of linear momentum.
LINEAR MOMENTUM AND AND IMPULSE IMPULSE 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.. (kg·m)/s 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 of N·s..
Principle of Impulse & Momentum
F m
dv dt
d ( mv ) dt
dG
dt
t 2
Fdt G
Fdt G
t 2
2
G1
G1
t 1
Fdt G t 1
Initial linear momentum + Linear impulse = Final momentum t 2
Fdt mv t 1
2
mv1
2
Principle of Impulse & Momentum The three components in x, y and z are t 2
F dt G x
2 x
G1 x mv 2 x mv 1 x
t 1 t 2
F dt G y
2 y
G1 y mv 2 y mv 1 y
t 1 t 2
F dt G z
t 1
2 z
G1 z mv 2 z mv 1 z
Principle of Impulse & Momentum Initial linear momentum + Linear impulse = Final momentum t 2
mv 1 x
F dt mv x
2 x
t 1 t 2
mv 1 y
F dt mv y
2 y
t 1 t 2
mv 1 z
F dt mv z
t 1
2 z
Principle of Impulse & Momentum t 2
Fdt mv
2
mv1
t 1
• 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.
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 • 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. • 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.
EXAMPLE 1 The 100-kg stone is originally at rest on the smooth horizontally surface. If a towing force of 200 N, acting at an angle of 45°, is applied to the stone for 10 s, determine the final velocity and the normal force which the surface exerts on the stone during the time interval.
EXAMPLE 1 Free Body Diagram
Impulse
I F c ( t 2 t 1 )
Velocity: horizontal and positive when pointing to the right
EXAMPLE 1
EXAMPLE 1
EXAMPLE 2 The 250-N crate is acted upon by a force having a variable magnitude P = (100t) N. Determine the crate’s velocity 2 s after P has been applied. The initial velocity is v1 = 1 m/s down the plane, and the coefficient of kinetic friction between the crate and the plane is μk = 0.3.
EXAMPLE 2 Free Body Diagram Impulse --- For varying force P: integrating P = 100t over the 2-s time interval. ---For constant forces (Weight, normal force and frictional force):
I F c ( t 2 t 1 )
EXAMPLE 2
EXAMPLE 2
EXAMPLE 3 Given: 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 3 Plan: 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: W dt 0 mv1
=
+ mvO = 0
F dt
40°
N dt 0
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).
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:
Conservation of Linear Momentum Considering two particles interacting during a time interval t , if the interaction forces are F one particle.
G1 Ft
Then it will be F
for the other particle.
G2 Ft
Conservation of Linear Momentum Therefore the total change in linear momentum for the system is
Ft ( Ft ) 0 G 0 or
G1 G2
This is known as the principle of conservation of linear momentum.
Principle of Linear Impulse & Momentum for a System of Particles For the system of particles shown, the internal forces f i between 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.
F i
mi a i
mi vi
mi
dvi dt
Principle of Linear Impulse & Momentum for a System of Particles The linear impulse and momentum equation for this system only includes the impulse of external forces.
t 2
m v F dt m v i i1
i
t 1
i i2
EXAMPLE 4 Given:Two rail cars with masses of m A = 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 when two bodies collide with each other during a very short period of time, causing relatively large (impulsive) forces to be exerted between the bodies.
Impact The line of impact passes through the mass centers of the particles. Central impact: the direction of motion of the mass centers of the two colliding particles is along the line of impact. Oblique impact: one or both of the particles is moving at an angle with the line of impact.
Central Impact • Consider two smooth particles A and B with, the initial momentum as shown • Provided ( v A ) 1 ( v B ) 1 , collision will Eventually occur.
Central Impact The particles are deformable or non-rigid during the collision. • An equal but opposite deformation impulse ∫ Pdt is 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 (the particles will either return to their original shape or remain permanently deformed). • The equal but opposite restitution impulse ∫R dt pushes the particle apart from one another.
Central Impact Just after the separation the particles will have the final momentum, where ( v B ) 2 ( v A ) 2
Central Impact Conservation of linear momentum
m A(v A )1 m B (v B )1 m A(v A )2 m B (v B )2 With two unknown velocities
(v A )2
and
(v B )2
Central Impact The coefficient of restitution.
e
(v B )2 (v A )2 (v A )1 (v B )1
(v B )2 (v A )2 Relative velocity just after impact
(v A )1 (v B )1 Relative velocity just before impact
Central Impact Coefficient of restitution e has a value between 0and 1 depending on the material property of the particles
Procedure for Analysis-Central Impact Find the final velocities just after 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 relative velocities before and just after collision.
EXAMPLE 5 The bag A, having a mass of 6 kg is released from rest at the position θ = 0°. After falling to θ = 90°, is strikes an 18 kg box B. If the coefficient of restitution between the bag and the box is e = 0.5, determine the velocities of the bag and box just after impact and the loss of energy during collision.
Solution
Coefficient of Restitution
Oblique Impact Particles 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
(v A )2 (v B )2 2 2 or (v Ax )2 (v Ay )2 (v Bx )2 (v By )2
Procedure for Analysis-Oblique Impact Choose the line of impact as the
x
axis.
The impulsive forces of deformation and restitution act only in the x direction. Resolving the velocities into components along x and y axes.
(v Ax )2 (v Ay )2 (v Bx )2 (v By )2
Procedure for Analysis-Oblique Impact Conservation of momentum along
(v ) (v ) x 1
Coefficient of restitution
e
(v Bx )2 (v Ax )2 (v Ax )1 (v Bx )1
x 2
x
axis.
Procedure for Analysis-Oblique Impact Conservation of momentum along impulse on A and B)
m A (v Ay )1 m A (v Ay )2 m B (v By )1 m B (v By )2
y
axis (no
EXAMPLE 6 Two smooth disks A and B, having mass of 1 kg and 2 kg respectively, collide with the velocities shown. If the coefficient of restitution for the disks is e = 0.75, determine the x and y components of the final velocity of each disk just after collision.
Solution Resolving each of the initial velocities into x and y components, we have
The four unknown velocity components after collision are assumed to act in the positive directions. Conservation of “ x” Momentum
Coefficient of ( x) Restitution. Both disks are assumed to have components of velocity in the +x direction after collision
