Momentum and Its Relation to Force

Transcription

1 Linear Momentum

2 Momentum and Its Relation to Force Momentum is a vector symbolized by the symbol p, and is defined as: It is a vector and has units of: (kg m/s) or (Ns) The rate of change of momentum is equal to the average net force: This can be shown using Newton s second law. F Net = ma = m Δ v Δt = m v f v i Δt = m v f m " v i Δt F Net = Δ p Δt F Net Δt = m v f m " v i = p f p i = Δ p F Net Δt = Δ p or F Net = Δ p Δt

3 Example 1: If a 1200 kg car has a momentum of 1.50x10 4 kg/s North, what is its velocity? Answer:12.5 m/s North Solution: p=mυ υ = p/m = (1.50x10 4 )/1200 = 12.5 m/s North The momentum vector has the same direction as the velocity vector.

4 Impluse is defined as an object s change in momentum (Δp). Impulse must also be a vector. On slide#2 it was shown how an object s change in momentum was related to force: F Net Δt = Δ p where Δ p = p f p i Example 2: (1-d vector problem) An 5.00 kg object was travelling to the right with a speed of 3.00 m/s. In a time of 1.20 s it slowed to speed of 1.00 m/s travelling in the same direction. a) What was the object s change in momentum? b) What average net force acted on the object?

5 Solution example 2: a) Sketch a diagram first, and define a positive direction : +X 3 m/s 1.00 m/s Δ p = p f p i Before After = (5)(1) - (5)(3)= -10 kg m/s b) F Net = Δ p Δt = -10/1.2= N

6 Example 3: A 5.00 kg object travelling to the right at 6.00 m/s collided with a wall and bounced back at 4.00 m/s in the opposite direction. a) What was the object s impulse? b) If the wall exerted an average force of 120 N on the object in the collision, for how long was the object in contact with the wall?

7 Example 3 Solution(1-d vector problem) : a) Before After 6.00 m/s 4.00 m/s +X Δ p = p f p i b) Δt = Δ p F Net = (5)(-4) (5)(6) = Ns = (-50)/(-120) = 0.42 s Recall that momentum units can be stated Ns or Kg m/s. They are both part of the mks system.

8 Examining momentum in 2- dimensions does not pose any difficulties to someone who understands 2-d vector analysis. The principles (definitions) of momentum in 1-d and 2-d are the same. Example 4 (2d vector problem) : A 5.00 Kg mass had an initial velocity of 3.00 m/s north. In a time of 2.43 s its velocity changed to 4.00 m/s East. a) What was the impulse of the mass? b) What average force acted on the mass over its momentum change?

9 Solution Example 4: a) p i = (5)(3) = 15 Ns north Before 3 m/s After p f = (5)(4) = 20 Ns East 4 m/s Δ p = p f p Use components or head to i tail method to solve for vector Δp Δ p = p f + ( p i ) p f = 20 Ns Δp = (15) 2 + (20) 2 = 25.0 Ns - p i = 15 Ns Θ= tan -1 ( 15/20)= S of E Δp =? b) F Net = Δ p Δt = 25/2.43 = 10.3 N S of E

10 Example 5 (2d vector problem) : An airplane (m= 3.50x10 5 kg) had an initial velocity of 400 km/h north. A net force of 2.48x10 5 N 71 0 E of S, acted on the plane for a time of 3.00 minutes. What was the final velocity (in km/h) of the plane? Find Impulse 1 st. One could solve using adding vectors head to tail but in this case solve using vector components for practice.

11 Solution Example 5: Δ p = p f p i p f = Δ p + p i Pi = 400 Km/h = 3.889x10 7 Ns North ΔP =FΔt (2.48x10 5 )[(3)(60)] = 4.464x10 7 Ns 71 0 E of S Using components: p fx = Δ p x + p ix = (4.464x10 7 )cos(19 0 ) + 0 = x10 7 Ns p fy = Δ p y + p iy = ( x10 7 )sin(19 0 ) x10 7 = x10 7 Ns p f = p 2 fx + p 2 fy = ( x 10 7 ) 2 + ( x 10 7 ) 2 = x10 7 Ns N of E p f p fx p fy Θ = tan -1 (pfy/pfx) = υ = p m = (4.873 x107 )/(3.50x10 5 )= 139 m/s = 501 km/h N of E Retry this problem by adding vectors head to tail and solving for P f using cosine and sine laws.

12 Collisions and Impulse During a collision, objects are deformed due to the large forces involved. In a collision there is an average net force that acts on the object. There is also an equal and opposite force acting on the other surface (Newton s 3 rd Law).

13 The graphs below show a typical collision (usually some sort of bell curve). The force of contact (reaction force) versus the time of contact. The area under the curve equals the impulse (recall Δp=FΔt), where F is the average net force.

14 Example 6: A kg mass approached a wall with a speed of 2.00 m/s. If the graph (triangle) below shows the magnitude of force that the mass experienced while in contact with the wall, determine the speed it left the wall with. Depending on the definition of + direction the impluse (area) could be defined by a + or - value. Be careful. Sketch a diagram and define direction. F(N) (50 ms, 35 N) 30 Area of triangle =0.5 (base)(height) F Net Δt = Δ p = " p f p i t (ms)

15 Example 6 solution: It is an arbitrary decision to define a + direction. Let the + direction be to the right (same direction as initial velocity). F Net Δt = Δ p = " p f p i Before x+ 2 m/s After F W A L L One can use the area under the curve to obtain the magnitude of the impulse. Area = (0.5)( 100 x10-3 s)(35 N) = 1.75 Ns One can see that the direction of the impulse (same direction as average net force on object) is to the left, so it should take on a - value. F Net Δt = Δ p = " p f p i = (0.5)υ (0.5)(2) υ= m/s speed = 1.50 m/s

16 The impulse tells us that we can get the same change in momentum with a large force acting for a short time, or a small force acting for a longer time. This is why you should bend your knees when you land; why airbags work; and why landing on a pillow hurts less than landing on concrete. There are of course less severe examples to test these concepts.

17 The Law of Conservation of Momentum During a collision, measurements show that the total momentum does not change. The total momentum before collision equals the total momentum after collision. p Total Before = p Total After F F During the collision there are equal and opposite forces. This means the impulse of each of the objects are equal in magnitude but in opposite directions. The sum of both impulses is equal to zero meaning there is no net change in the total momentum of the system (system = both masses). This indicates that the total momentum of the system should remain constant. Momentum conservation follows from Newton s 3 rd law.

18 The total momentum of an isolated system of objects remains constant. The total momentum of the system before collision (explosion) is equal to the total momentum after collision. It should be noted that even in an explosion, where potential energy is changed to kinetic, newton s 3 rd law applies.

19 Momentum conservation works for a rocket as long as we consider the rocket and its fuel to be one system, and account for the mass loss of the rocket.

20 Momentum is conserved in all collisions (explosions). Collisions in which kinetic energy is conserved as well are called elastic collisions, and those in which it is not are called inelastic. Often to emphasize that a collision is elastic, one calls it a perfectly elastic collision. If the two objects stick together after collision it is called a completely inelastic collision.

21 Most collisions are inelastic. With inelastic collisions, some of the initial kinetic energy is lost to other forms of energy such as thermal or sound energy. See the example on the next slide.

22 Cars and Cliffs

23 1-D conservation of Momentum Examples When solving any momentum problem get into the habit of drawing a diagram of before and after. Define your directions. State the relevant law or relationship. Obtain equation(s) and solve. Example 7: A 4.00 kg object was travelling east at 2.00 m/s and collided with a 3.00 kg object travelling west at 1.00 m/s. After the collision the 3.00 kg object had a speed of 1.00 m/s in the opposite direction (east) a) What was the velocity of the 4.00 kg mass after the collision? b) Was this a perfectly elastic collision? Compare the total kinetic energy (of both objects) before and after the collision. c) What kind of collision was it?

24 Example 7 solution: First sketch a diagram and define a positive direction. x+ 2 m/s 1 m/s 1 m/s 4 kg 3 kg 4 kg 3 kg a) p Total Before = p Total After Before (4)(2) + (3)(-1) = 4υ + (3)(1) υ= 0.50 m/s After b) E K total before = (0.5)(4)(2) 2 + (0.5)(3)(1) 2 = 9.50 J E K total after = (0.5)(4)(0.5) 2 + (0.5)(3)(1) 2 = 2.00 J There was 7.50 J lost it is not a perfectly elastic collision c) It was an inelastic collision since some kinetic energy changed form. Note that the kinetic energy ( always be a positive number. 1 2 mυ 2 )of any object must

25 Example 8: A 10.0 kg object was initially travelling West (to the left) at 15 m/s and exploded into two pieces. A 4.00 kg piece after the explosion was travelled 7.50 m/s East (to the right). a) What was the velocity of the other piece after the explosion? b) How could one verify that this was indeed an explosion and not a typical collision?

26 Solution Example 8: Draw a diagram and define a direction. I labeled the unknown velocity υ in the direction of +x. One would expect that υ must be in the other direction, so a negative value is expected. x+ a) p Total Before = p Total After 15 m/s 10.0 kg 4 kg Before 7.50 m/s (10)(-15) = (6)υ + (4)(7.5) υ= -30 m/s, or 30 m/s left. One must clearly indicate a direction when stating momentum (unless stated otherwise). =? After b) One can determine that it must be an explosion by calculating the total kinetic energy after and before the explosion. If the total kinetic energy is greater after the explosion than before some form of potential energy must have been converted into kinetic energy (i.e in an explosion).

27 Example 9: A 4.00 kg mass travelling at 5.00 m/s to the right had a completely inelastic collision with a 7.00 kg mass travelling at 4.00 m/s in the opposite direction. What was the velocity of the masses after the collision?

28 Example 9 solution: If it is completely inelastic then the two objects join or stick together. One can treat it the masses after collision as a combined mass. p Total Before = p Total After 5 m/s 4 m/s 4 kg 7 kg Before x+ 11 kg After (4)(5) + (7)(-4) = 11υ υ = m/s or 0.73 m/s left

29 Here we have two objects colliding elastically. We know the masses and the initial speeds. Since both momentum and kinetic energy are conserved, we can write two equations. This allows us to solve for the two unknown final speeds.

30 Example 10: A 2.00 kg object was travelling at 4.00 m/s to the right and collided perfectly elastically with a 3.00 kg object travelling to the left at 1.50 m/s. Set up the two equations necessary to solve for the unknown velocities after the collision. Always draw a before and after collision diagram.

31 Example 10 solution: x+ Before Collision After Collision 2 kg 3 kg kg 3 kg 4 m/s 1.5 m/s Momentum is conserved: p Total Before = p Total After (2)(4)+(3)(-1.5) = 2υ 1 + 3υ 2 Kinetic energy is conserved: 3.5 = 2υ 1 + 3υ 2 Total kinetic energy before collision = Total kinetic energy after collision (0.5)(2)(4) 2 + (0.5)(3)(1.5) = (0.5)(2) υ 1 + (0.5)(3) υ = υ υ 2 2 For a challenge solve the system of equations to find the unknown velocities after the collision.

32 Additional Momentum/Energy Problems: Example 11(harder): Two stationary masses (one mass is double the mass of the other) on a frictionless have a compressed spring sandwiched between them. The spring is released and the smaller mass (m) had a velocity υ after the release of the spring. What was the amount of stored energy (in terms of m and υ) in the spring if 80 % of this energy went into the movement of the masses? Use momentum 1 st. m Before compressed spring 2m frictionless ground m m After frictionless ground 2m '=?

33 Example 11 solution: p Total Before = p Total After + 0 = -mυ + 2mυ υ = 0.5υ The total kinetic energy of the objects after the spring released its energy is : 0.5mυ (2m)(0.5υ) 2 = (0.75)mυ 2 Only 80% of the store energy went into the movement of the objects, therefore the energy E initially stored in the spring is: (0.80)E =(0.75) mυ 2 E= (0.938)mυ 2

34 Example 12: A gram bullet travelling horizontally at 285 m/s struck a stationary 1.35 Kg pendulum and had a speed of 80.0 m/s after it passed through the pendulum mass. To what vertical height did the pendulum swing to? Start with momentum (collision) then use energy conservation. Before After h=? 285 m/s 80 m/s

35 Example 12 solution: ptotal Before = p Total After + (0.015)(285) = (0.015)(80) + (1.35)υ υ = m/s Conservation of mechanical energy: 1 2 mυ 2 = mgh 1 2 υ 2 = gh h = (2.778) 2 / [2(9.80)] = m

36 Conservation of energy and momentum can also be used to analyze collisions in two or three dimensions, but unless the situation is very simple, the math quickly becomes unwieldy. Here, a moving object collides with an object initially at rest. Knowing the masses and initial velocities is not enough; we need to know the angles as well in order to find the final velocities.

37 Example 13 (2-d): A 2.00 kg object was initially travelling east at 10.0 m/s and collided with a stationary 8.00 kg mass. After the collision the 2.00 kg mass had velocity of 5.00 m/s 30 0 N of E. What was the velocity of the 8.00 Kg mass after the collision? Steps to follow: 1) Sketch a diagram (before and after collision). 2) Define your momentum vectors 3) State a vector relationship dictated by conservation of momentum. 4) Draw the vector diagram according to how the three momentum vectors are related. 5) Use cosine and sine laws to find the unknown momentum vector.6) Determine velocity. One could also use components to do this problem. Make sure that you can apply components to this problem.

38 Example 13 solution: p 1 = 20 Ns East p 2 p 3 =? = 10 Ns 30 0 N of E Conservation of momentum: Vector relationship: W N S 2 kg E Before 10 m/s p 1 p Total Before = p Total After p 1 = p 2 + p 3 v= 0 8 kg After p 2 5 m/s 2 kg kg p 3? Vector diagram: p 2 = 10 Ns p 3 = (10)(20)cos30 0 = Ns p 3 =? 30 0 p 1 = 20 Ns υ = p m = /8 = 1.55 m/s sinθ 10 = sin In summary, Θ= υ = 1.55 m/s S of E

39 Often the choice to use components or not, is just an arbitrary decision. There are some circumstances however where components may be more convenient. The case where there are more than 3 momentum vectors involved makes the choice much easier. In the following example, using components is the way to go. Example 14 (2-d): A 10.0 kg object with some unknown velocity exploded into 3 pieces. After the explosion the 2.00 kg piece travelled north at 4.00 m/s, the 5.00 kg piece travelled East at 6.00 m/s, and the remaining piece travelled at 8.00 m/s 30 0 E of S. What was the velocity of the 10.0 kg piece before the collision? Steps to follow: 1) Sketch a diagram (before and after collision). 2) Define your momentum vectors 3) State a vector relationship dictated by conservation of momentum. 4) Determine the x and y components of all vectors. 6) Using the vector relationship solve for the unknown x and y components. 7) Determine the magnitude and direction of the unknown momentum vector.

40 Example 14 Solution: p 1 =? p 2 = 8 Ns N p 3 = 30 Ns E P 4 = 24 Ns 30 0 E of S p 2x = 0, p 2y = 8 Ns p 3x = 30 Ns, p 3y = 0 P 4x = 24(cos60) Ns p 4y = -(24)(sin60) Ns Before 10 kg p Total Before = p p1 Total After = p 2 + p 3 + p 4 p 1x = p 2 x + p 3x + p 4 x +y +x p 1 =? = (cos60) = 42 Ns After p 2 4 m/s 2 kg 5 kg kg 8 m/s p 4 6 m/s p 3 p 1y = p 2 y + p 3y + p 4 y = (sin60) = Ns 42 Ns p Ns p 1 = p 2 1x + p 1y " θ = tan % $ # 42 & 2 = Ns υ = p 1 m = 43.9/10= 4.39 m/s ' = υ = 4.39 m/s S of E