1. Prepared by Dr. Hassan Fadag.
Lecture X
Impulse & Momentum

2. i) Linear Impulse & Linear Momentum
From Newton’s 2nd Law:
SF = m a
= m v.= d/dt (m v)
The term m v is known as the linear momentum, and it is abbreviated by G.
SF = G (1)
This formula states that the resultant of all forces acting on a particle equals its time rate of change of linear momentum.
The SI unit of the linear momentum is N.s.
Note that the rate of change of linear momentum G. is a vector quantity, and its direction coincides with the direction of the resultant forces, i.e. coincides with the direction of the acceleration. Where the direction of G coincides with the direction of the velocity.
In rectangular coordinates, the three scalar components of formula (1) are:
SFx = Gx SFy = Gy SFz = Gz.
Note: the momentum formula is valid as long as the mass m is not changing.

3. Linear Impulse & Linear Momentum – Cont.
Force, F F2
From formula (1),
SF = G.
SF = d/dt (G)
SF dt = d (G)
By the integration of both sides,
The product of force and time, represented by the term , is known as the linear impulse.
The formula (2) states that the total linear impulse on m equals to the change in the linear momentum of m.
The SI unit of the linear impulse is N.s.
In rectangular coordinates, the magnitudes versions of (2) are: F1 t1
Time, t t2
Note: If the force, which is acting on a particle, varies with time, the impulse will be the integration of this force w.r.t. time or the area under the above curve. or
(2)
Conservation of Linear Momentum
When the sum of all forces acting on a particle is zero, i.e SF = 0, thus,
DG = 0 or G1 = G2

4. ii) Angular Impulse & Angular Momentum
The angular momentum (HO) is the moment of the linear momentum vector mv at point P about the origin O, and it is given by:
(3)
So, the angular momentum is a vector perpendicular to plane A. In rectangular coordinates, the three scalar components of formula (3) are:
So from figure (b) , the magnitude of the angular momentum is simply the linear momentum mv times the moment arm r sinq
or:
The SI unit of the angular momentum is N.m.s.

5. Angular Impulse & Angular Momentum – Cont.
Rate of change of Angular Momentum
From formula (1), SF = G. = m v. , by taking the moment Mo of particle P about origin O for both sides of the formula we get:
By differentiating formula (3), we get:
This makes: (4)
i.e. the moment about the fixed point O of all forces acting on m equals the time rate of change of angular momentum
of m about O. So the components for formula (4):
The form is another form of formula (4),
which leads to By integrating both sides:
Note: The term v x mv is zero since the cross product of parallel vectors is identically zero. or
(5)

6. Angular Impulse & Angular Momentum – Cont.
The product of moment and time is known as the angular impulse.
The formula (5) states that the total angular impulse on m about the fixed point O equals the corresponding change in angular momentum of m about O.
The SI unit of the angular impulse is N.m.s.
The x-component for formula (5):
Similar expressions exit for y and z components.
For plane motion applications (2D problems) in xy-plane; the scalar form of formula (5) is: or
Conservation of Angular Momentum
When the resultant moment about a fixed point O of all forces acting on a particle is zero during an interval of time, i.e SMo = 0, thus,
DHo = 0 or (Ho)1 = (Ho)2 or

7. Impulse & Momentum Exercises

8. Exercise # 1
A particle with a mass of 0.5 kg has a velocity of 10 m/s in the x-direction at time t = 0. Forces F1 and F2 act on the particle, and their magnitudes change with time according to the graphical schedule shown. Determine the velocity v2 of the particle at the end of the 3-s interval. The motion occurs in the horizontal x-y plane.

9. Exercise # 2
If the jets exert a vertical thrust of T = (500t3/2) N, where t is in seconds, determine the man’s speed when t = 3 s. The total mass of the man and the jet suit is 100 kg. Neglect the loss of mass due to the fuel consumed during the lift which begins from rest on the ground.

10. Exercise # 3
The 1500-kg car has a velocity of 30 km/h up the 10-percent grade when the driver applies more power for 8 s to bring the car up to a speed of 60 km/h. Calculate the time average F of the total force tangent to the road exerted on the tires during the 8 s. Treat the car as a particle and neglect air resistance.

11. Exercise # 4
The 2.5-Mg van is traveling with a speed of 100 km/h when the brakes are applied and all four wheels lock. If the speed decreases to 40 km/h in 5 s, determine the coefficient of kinetic friction between the tires and the road.

12. Exercise # 5
The 5-kg block is moving downward at v1 = 2 m/s when it is 8 m from the sandy surface. Determine the impulse of the sand on the block necessary to stop its motion. Neglect the distance the block dents into the sand and assume the block does not rebound. Neglect the weight of the block during the impact with the sand.

13. Exercise # 6
The crate B and cylinder A have a mass of 200 kg and 75 kg, respectively. If the system is released from rest, determine the speed of the crate and cylinder when t = 3 s. Neglect the mass of the pulleys.

14. Exercise # 7
The 50-g bullet traveling at 600 m/s strikes the 4-kg block centrally and is
embedded within it. If the block slides on a smooth horizontal plane with a velocity of 12 m/s in the direction shown prior to impact, determine the velocity v2 of the block and embedded bullet immediately after impact.

15. Exercise # 8
A small sphere has the position and velocity indicated in the figure and is acted upon by the force F. Determine the angular momentum HO about point O and the time derivative H˙o.

16. Exercise # 9
The 3-kg sphere moves in the x-y plane and has the indicated velocity at a particular instant. Deter- mine its (a) linear momentum, (b) angular momentum about point O, and (c) kinetic energy.

17. Exercise # 10
The 6-kg sphere and 4-kg block (shown in section) are secured to the arm of negligible mass which rotates in the vertical plane about a horizontal axis at O. The 2-kg plug is released from rest at A and falls into the recess in the block when the arm has reached the horizontal position. An instant before engagement, the arm has an angular velocity wo = 2 rad/s. Determine the angular velocity w of the arm immediately after the plug has wedged itself in the block.

18. Prepared by Dr. Hassan Fadag.
Special Applications
a) Impact b) Relative Motion

19. a) Impact
Impact: is the collision between two bodies. It leads to the generation of relatively large contact forces that act over a very short interval of time. There are two main types of impact, they are: a) direct central impact and b) oblique central impact.
a) Direct central impact: a collinear motion of two spheres of masses m1 and m2, traveling with velocities v1 and v2. If v1 > v2, collision will occur with contact forces directed along the line of centers of the two spheres. Note that the sum of forces are zero or relatively small (produces negligible impulses), thus, F1 F2

20. By eliminating vo in both equation
Impact – Cont.
Coefficient of restitution (e): in the direct impact formula, there are two unknowns An additional relationship is needed to find the these final velocities. This relationship must reflect the capacity of the contacting bodies to recover from the impact and can be expressed in by the ratio e of the magnitude of the restoration impulse to the magnitude of the deformation impulse. This ratio is known as the coefficient of restitution.
For particle 1
By eliminating vo in both equation
For particle 2

21. Impact – Cont.
b) Oblique central impact: the initial and final velocities are not parallel. The directions of the velocity vectors are measured from the direction tangent to the contacting surfaces.
However, there are four unknowns; they are To find them, the following four formulas will be used:
(6)
(9)
(7)
Note: There is no impulse on either particle in the t-direction
(8)

22. Exercise # 11
The ram of a pile driver has a mass of 800 kg and is released from rest 2 m above the top of the 2400-kg pile. If the ram rebounds to a height of 0.1 m after impact with the pile, calculate (a) the velocity vp' of the pile immediately after impact, (b) the coefficient of restitution e, and (c) the percentage loss of energy due to the impact.

23. Exercise # 12
The 20-kg package has a speed of 1.5 m/s when it is delivered to the smooth ramp. After sliding down the ramp it lands onto a 10-kg cart as shown. Determine the speed of the cart and package after the package stops sliding on the cart.

24. Exercise # 13
The bus B has weight WB (75 KN) and is traveling to the right at speed VB (2.5 m/s). Meanwhile car A of weight WA (15 KN) is traveling at speed VA (2 m/s) to the left. If the vehicles crash head-on and become entangled, determine their common velocity just after the collision. Assume that the vehicles are free to roll during collision.

25. Exercise # 14
Compute the final velocities v1' and v2' after collision of the two cylinders which slide on the smooth horizontal shaft. The coefficient of restitution is e = 0.6.

26. Exercise # 15
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.

27. Exercise # 16
The cue ball A is given an initial velocity (vA)1 = 5 m/s. If it makes a direct collision with ball B (e = 0.8), determine the velocity of B and the angle q just after it rebounds from the cushion at C (e ’ = 0.6). Each ball has a mass of 0.4 kg. Neglect the size of each ball.

28. Note: when the system is accelerating :
b) Relative Motion
The acceleration of A as observed from or relative to x-y-z is arel = aA/B = ¨rA/B, and by the relative-motion principle, the absolute acceleration of A is:
Thus, Newton’s 2nd law becomes:
(10)
D’Alembert’s Principle:
When we observe the particle from a moving system x-y-z attached to the particle, Fig.b, the particle necessarily appears to be at rest or in equilibrium in x-y-z. Thus, the observer who is accelerating with x-y-z concludes that a force -ma acts on the particle to balance ΣF. This point of view, which allows the treatment of a dynamics problem by the methods of statics, was an outgrowth of the work of D’Alembert.
Note: when the system is accelerating :
SF = marel

29. Relative Motion – Cont.
The fictitious force –ma is known as the inertia force, and the artificial state of equilibrium created is known as dynamic equilibrium.
Constant-Velocity, Nonrotating Systems:
In a moving system x-y-z , if vB is constant, then aB = 0. Thus, , i.e.:
Such a system is known as an inertial system or as a Newtonian frame of reference.
Work-energy, Impulse-Momentum relations holds for the case of constant velocity.
and

30. Exercise # 17
The flatbed truck is traveling at the constant speed of 60 km/h up the 15-percent grade when the 100- kg crate which it carries is given a shove which im-parts to it an initial relative velocity x˙ = 3 m/s toward the rear of the truck. If the crate slides a distance x = 2 m measured on the truck bed before coming to rest on the bed, compute the coefficient of kinetic friction mk between the crate and the truck bed.

31. Exercise # 18
The cart with attached x-y axes moves with an absolute speed v = 2 m/s to the right. Simultaneously, the light arm of length l = 0.5 m rotates about point B of the cart with angular velocity q˙= 2 rad/s. The mass of the sphere is m = 3 kg. Determine the following quantities for the sphere when q = 0: G, Grel, T, Trel, HO, (HB)rel where the subscript “rel” indicates measurement relative to the x-y axes. Point O is an inertially fixed point coincident with point B at the instant under consideration.

32. Exercise # 19
The launch catapult of the aircraft carrier gives the 7-Mg jet airplane a constant acceleration and launches the airplane in a distance of 100 m measured along the angled takeoff ramp. The carrier is moving at a steady speed vC = 16 m/s. If an absolute aircraft speed of 90 m/s is desired for takeoff, determine the net force F supplied by the catapult and the aircraft engines.