Ivan Ker B

Ivan ker B. saligumba

BSIT-2A

physics 101

Linear Momentum Linear momentum of an object is defined as the product of the object’s mass and its velocity. Momentum is generally represented by the letter "p." Since velocity is a vector quantity, momentum is also a vector, and has the same direction as velocity. The SI unit for momentum is kilograms times meters per second, or k⋅ m/s. The concept of momentum lets us generalize Newton’s second law of motion. Recall Newton’s second law: an object’s change of motion is proportional to both the magnitude of the applied force and the object’s mass. Equation 2.1, F = ma, was a translation of this law that was only valid for special cases: those in which the object’s mass was constant. However, utilizing the

definition of momentum, the general equation is below: For a constant mass, D p = mD v. Thus,

, mv + mv = (m+m)v The linear momentum p of an object with mass m and velocity v is defined as

From this definition it is clear that the unit of momentum is (kg m/s) or (N s). Since this momentum is related to the linear motion of the object, it is called linear momentum. In Chapter 11 we will be discussing angular momentum which is the momentum related to the angular motion of the object. Under certain circumstances the linear momentum of a system is conserved. The linear momentum of a particle is related to the net force acting on that object:

The rate of change of linear momentum of a particle is equal to the net force acting on the object, and is pointed in the direction of the force. If the net force acting on an object is zero, its linear momentum is constant (conservation of linear momentum). The total linear momentum p of a system of particles is defined as the vector sum of the individual linear momenta

This expression can be rewritten as

where M is the total mass of the system. We conclude that

" The linear momentum of a system of particles is equal to the product of the total mass M of the system and the velocity of the center of mass. " If we differentiate linear momentum of the center-of-mass with respect to time we obtain

This expression shows that if the net external force acting on a system of particles is zero (Fext = 0 N), the linear momentum of the system is conserved.

Problem: 1) An atomic nucleus intially moving @ 420 m/s emits an alpha particle in the direction of it's velocity, and the remaining nucleus slows to 350 m/s. If the alpha particle has a mass of 4.0 u and the original nucleus has a mass of 222 u what speed does the alpha particle have when it is emitted. mv + mv = (m+m)v (222 u)(420 m/s) = (218 u)(350 m/s) + (4.0 u)(v) v = 4.2 x 10^3 m/s Answer: 4.2 x 10^3 m/s 2.) A 12,600 kg railroad car travels alone on a level frictionless track with a constant speed of 18 m/s. A 5350 kg load, initially at rest, is dropped on to the car. What will be the car's new speed? mv + mv = (m+m)v (12600 kg)(18m/s) + (5350 kg)(0 m/s) = (17950 kg)(v) v= 12.6m/s Answer: 12.6 m/s

Elastic Collision elastic collision is a collision in which the total kinetic energy of the colliding bodies after collision is equal to their total kinetic energy before collision. Elastic collisions occur only if there is no net conversion of kinetic energy into other forms. During the collision kinetic energy is first converted to potential energy associated with a repulsive force between the particles (when the particles move against this force, i.e. the angle between the force and the relative velocity is obtuse), then this potential energy is converted back to kinetic energy (when the particles move with this force, i.e. the angle between the force and the relative velocity is acute).The collisions of atoms are elastic collisions (Rutherford backscattering is one example).

Inelastic Collision An inelastic collision is a collision in which kinetic energy is not conserved (see elastic collision). In collisions of macroscopic bodies, some kinetic energy is turned into vibrational energy of the atoms, causing a heating effect, and the bodies are deformed.

The formulas for the velocities after a one-dimensional collision are:

where V1f is the final velocity of the first object after impact V2f is the final velocity of the second object after impact V1 is the initial velocity of the first object before impact V2 is the initial velocity of the second object before impact M1 is the mass of the first object M2 is the mass of the second object CR is the coefficient of restitution; if it is 1 we have an elastic collision; if it is 0 we have a perfectly inelastic collision, see below. In a center of momentum frame the formulas reduce to: V1f = − CRV1 V2f = − CRV2 For two- and three-dimensional collisions the velocities in these formulas are the components perpendicular to the tangent line/plane at the point of contact.

Problems:

1) Two blocks are travelling toward each other. The first has a speed of 10 cm/sec and the second a speed of 60 cm/sec. After the collision the second is observed to be travelling with a speed of 20 cm/sec in a direction opposite to its initial velocity. If the weight of the first block is twice that of the second, determine: (a) the velocity of the first block after collision; (b) whether the collision was elastic or inelastic. Solution: We have a collision problem in 1–dimension. We draw both 'before' and 'after' pictures and select a coordinate system as shown.

B e fo r e A

A fte r

vA

i

vB

i

B

A

x

vA

f

B

vB

f

x

Since the surface is frictionless, and since no work is performed by either mg or the normal, then the net force acting on the system is 0, and we have conservation of linear momentum: p1i ⊕ p2i

=

p1i ⊕ p1f

Thus adding the x–components we have: Since m1 = 2 m2 we find: Thus

2 v1f = – 60 and 08-2

m1v1i – m2v2i = m1 v1f + m2 v2f

2 v1i – v2i = 2 v1f + v2f  (2)(10) – (60) = 2 v1f + 20 v1f = – 30 cm/sec ('–' means to left). KEI = (1/2) m1 (v1I)2 + (1/2) m2 (v2I)2 . This gives:

The initial KE is given by:

= (1/2)(2 m2)(10)2 + (1/2) m2 (60)2 = (1/2)(200 + 3600) m2 = 1900 m2 KEf = (1/2) m1 (v1f)2 + (1/2) m2 (v2f)2 . This gives:

The final KE is:

= (1/2)(2 m2)(30)2 + (1/2) m2 (20)2 = (1/2)(1800 + 400) m2 = 1100 m2 Since KEf is not equal to KEI, the collision is inelastic. 2) A block of mass 200 g, sliding with a speed of 12 cm/sec on a smooth level surface, makes a head–on, elastic collision with a block of unknown mass, initially at rest. After the collision the velocity of the 200 g block is 4 cm/sec in the same direction as its initial velocity. Determine the mass of the 2nd block and its speed after the collision. Solution: We have a collision problem in 1–dimension. We draw both 'before' and 'after' pictures and select a coordinate system as shown. B e fo r e

A fte r

vA

i

A

B

A

vA

f

B

vB

f

x

x

Since the surface is frictionless, and since no work is performed by either mg or the normal, then the net force acting on the system is 0, and we have conservation of linear momentum: p1I ⊕ p2I

=

p1I ⊕ p1f

Thus adding the x–components we have:

m1v1I + 0 = m1 v1f + m2 v2f .

This gives: (200)(12) = (200)(4) + m2 v2f . Since we have 2 unknowns we look for an 'energy condition'. We are told that the collision is elastic. Hence, we also have conservation of KE and write: (1/2)m1(v1I)2 + (1/2)m2(v2I)2 = (1/2)m1(v1f)2 + (1/2)m2(v2f)2

We may cancel the (1/2) factors, and we obtain:

(200)(12)2 + 0 = (200)(4)2 + m2(v2f)2 .

Thus (m2 v2f) v2f = (200)(8) v2f = (200)(144) – (200)(16) or 8 v2f = 128. Hence: v2f = 16 cm/sec, and

m2 = (200)(8)/(16) = 100 gm .

3) A 2000 kg automobile going east on Chestnut Street at 60 mi/hr collides with a 4000 kg truck which is going south across Chestnut Street at 20 mi/hr. If they become coupled on collision, what is the magnitude and direction of their velocity immediately after colliding? Solution: We have a collision problem in 2–dimension. We draw both 'before' and 'after' pictures and select a coordinate system as shown. We have conservation of linear momentum: pAi ⊕ pB i = pA f ⊕ pB f .

vA

N i

E vB

N v

a fte r

f

i

E b e fo r e

θ

We have a completely inelastic collision problem. Applying Conservation of Linear Momentum we have: m1 v1i ⊕ m2 v2i = m1 v1f ⊕ m2 v2f We now rewrite this vector equation in terms of components (East & South). E comp: m1 v1i + 0 = (m1 + m2) vf cos θ S comp:

0 + m2 v2i = (m1 + m2) vf sin θ

Thus we have 2 equations in 2 unknowns. (Note that since cons. of linear momentum is an equi– dimensional equation, we do not need to convert units). E comp: (2000)(60) = (6000) vf cos θ

S comp: (4000)(20) = (6000) vf sin θ

Dividing the 'S' equation by the 'E' equation eliminates vf and yields: tan

= (sin θ )/(cos θ ) = (4000)(20)/(2000)(60) = 2/3 

θ

= 33.7o .

Then from the 'E' equation: vf = (2000)(60)/(6000) cos 33.7 = 24.04 mi/hr . 37

vA

A

4) Object A traveling with a speed of 25 m/sec collides with an identical object B which was initially at rest. After the collision, the two objects are moving as shown. A) Determine the speeds of the two objects after collision. b) Was the collision elastic or inelastic? (show work).

53

f

A fte r

B

vB

f

Solution: We have a collision problem in 2–dimensions. We draw both 'before' and 'after' pictures and select a coordinate system as shown. We have conservation of linear momentum: pAi ⊕ pB I = pA f ⊕ pB f . y A

vA

vA

A B e fo r e

i

B

For the y–components we have: (4/5) Hence

37

53

x

f

x

A fte r

B

vB

f

0 = mA vAf sin 37 - mB vB f sin 53  m vA f (3/5) = m vB f

vB f = (.6/.8) vA f = (3/4) vA f .

08-7

For the x–components we have: mA vo = mA vA f cos 37 + mB vB f cos 53  m vo = m vA f (4/5) + m vB f (3/5) Canceling the 'm' in each term and using vo = 25 m/sec. 25 = vA f (4/5) + (3/4) vA f (3/5) or 25 = vA f{(4/5) + (9/20)} = vA f = 20 m/sec

and

(25/20) v1f . Thus

vB f = 15 m/sec.

Note that the problem did not say whether the collision was elastic or inelastic. Since we have both final velocities then we can directly answer whether or not the collision was elastic. The initial KE is simply KEi = (1/2)m vo2 = (1/2) m (25)2 = (1/2)(625) m

The final KE would be: KEf = (1/2) mA (vA f)2 + (1/2) mB (vB f)2 . This gives: = (1/2) m (20)2 + (1/2) m (15)2 = (1/2) m { 400 + 225} = (1/2) m (625) . The collision is elastic.

TORQUE Torque is a measure of how much a force acting on an object causes that object to rotate. The object rotates about an axis, which we will call the pivot point, and will label 'O'. We will call the force 'F'. The distance from the pivot point to the point where the force acts is called the moment arm, and is denoted by 'r'. Note that this distance, 'r', is also a vector, and points from the axis of rotation to the point where the force acts. (Refer to Figure 1 for a pictoral representation of these definitions.)

Figure 1 Definitions

Torque is defined as = r x F = r F sin( ). In other words, torque is the cross product between the distance vector (the distance from the pivot point to the point where force is applied) and the force vector, 'a' being the angle between r and F. EXAMPLE PROBLEM ON TORQUE Problem 1 In a hurry to catch a cab, you rush through a frictionless swinging door and onto the sidewalk. The force you extered on the door was 50N, applied perpendicular to the plane of the door. The door is 1.0m wide. Assuming that you pushed the door at its edge, what was the torque on the swinging door (taking the hinge as the pivot point)? Solution

The pivot point is at the hinges of the door, opposite to where you were pushing the door. The force you used was 50N, at a distance 1.0m from the pivot point. You hit the door perpendicular to its plane, so the angle between the door and the direction of force was 90 degrees. Since

Figure 1 Diagram of Example Problem 1

= r x F = r F sin( ) then the torque on the door was: = (1.0m) (50N) sin(90)

= 50 N m

Problem 2 A 3.0kg mass is place 2.00m to the right of the pivot point of a see-saw. What is the the magnitude and the sign of the torque applied? This problem looks like the figure

The force exerted by the mass is due to gravity and is found from F=mg. The distance between the force and the pivot point is r=2.00m. We can find the magnitude of the torque by

If the mass is to the right of the pivot point, the rotation will be in a clockwise direction, so the torque is negative: . As always, note the use of significant figures. The distance r was given to three significant figures, but the mass (and therefore the force) is only known to two significant figures. Thus the torque must have only two significant figures.

Second Condition of Equilibrium An object in equilibrium does not move along a straight line -- it does not translate -- that means the sum of all the forces on it is zero. That was the first condition of equilibrium. But an object in equilibrium also does not rotate. That means the sum of all the rotational forces on it is also zero. The sum of all the torques on an object is equilibrium is zero. This is the Second Condition of Equilibrium. Tor ques that would rotate an object counter clockwise may be taken as positive and torques that would rotate an objectclockwise may be taken as negative. Then we can write this Second Condition of Equilibrium as

or we can calculate the sum of the clockwise torques and set them equal to the sum of the counterclockwise torques. Then we can write this Second Condition of Equilibrium as

Problem : 1) When only the front wheels of an automobile are run onto a platform scale the scale balances at 8 kN; when only the rea wheels are run onto the scale, it balances at 6 kN. What is the weight of the automobile and how far is its center of gravity behind the front axle? The distance between the axles is 2.8 m. Solution: We have a 'general' equilibrium problem since all forces in the problem do not act through a single point. This means that the problem cannot be solved using only the 1st condition of equilibrium (Fnet = 0). We must also employ the 2nd condition: ( Γ net = 0). In the 1st figure we draw the car with its front wheel on the balance. The balance reads 'Fcar on balance' = 8 kN. Hence, the force exerted on the car (N f) is also 8 kN.

N

2 .8 m r

N

x

f

S c a le

W

In the 2nd figure, with the rear wheels on the balance, we find that F2 = 6kN. Hence, the 1st condition of equilibrium gives us: Σ F hor: 0 = 0 ;

N

2 .8 m r

N

f

Σ Fver: N f + N r – W = 0

Thus the weight of the car is W = 14 kN. S c a le

x

W

The determination of the location of the center of gravity (point where W acts) involves distances & hence torques. We have indicated W acting at the CM (distance 'x' from rear axle). The 2nd condition of equilibrium can be applied about any point. We select point 'A' as the rear axle: Σ Γ

A

= Σ Γ

- Σ Γ

ckw

cckw

=

W (x) - N f (2.8) = 0

Inserting the know values for W & F2, we have: (8 kN)(2.8 m) =

(14 kN) x



x = 1.6 m

2) A 600 N bricklayer is 1.5 m from one end of a uniform scaffold 7 m long, weighing 800 N. A pile of bricks weighing 500 N is 3 m from the same end. If the scaffold is supported at the two ends, calculate the force on each end. Solution: We have a general equilibrium problem since all forces do not act through a common point. Since we must use the 2nd condition of equilibrium to solve this problem, We jump right in and calculate torques. Now there are 2 'unknown forces' T1 and T2 . Application of the torque condition about any point gives us a single scalar equation. Thus we need to choose points so that one of the unknown forces will be eliminated. If we choose point 'A' as the left end we will eliminate T2. Thus: Σ Γ N.

A

= Σ Γ

ckw

- Σ Γ

cckw

T T

1

2

W

3 .5 m

s

W

b

1 .5 m

7 m

W

m

1 .5 m

= W s (3.5) + Wb (4) + Wm (5.5) - T1 (7)

= (800)(3.5) + (500)(4) + (600)(5.5) – T1(7) = 0.

This gives:

We now choose to apply the 2nd condition about the right end:

T1 = (8100)/7 = 1157

Σ Γ

B

= Σ Γ

ckw

-Σ Γ

cckw

= T 2 (7) - W m (1.5) - W b (3) - W s (3.5)

= 7 T2 – (800)(3.5) – (500)(3) – (600)(1.5) = 0 .

Thus:

T2 = (5200)/7 = 743 N.

Now as a check in the problem we use the 1st condition of equilibrium. Σ Fup = T1 + T2 = 1157 + 743 = 1900 N. Σ Fdown = Wplank + Wman + Wbricks = 800 + 600 + 500 = 1900 N.

Hence our answers check!