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The Conservation of Mechanical Energy Example: A Daredevil Motorcyclist A motorcyclist is trying to leap across the canyon by driving horizontally off a cliff 38.0 m/s. Ignoring air resistance, find the speed with which the cycle strikes the ground on the other side. The Conservation of Mechanical Energy Ef = Eo 2 f 2 o mgh f + mv = mgho + mv 1 2 1 2 gh f + 12 v 2f = gho + 12 vo2 The Conservation of Mechanical Energy gh f + 12 v 2f = gho + 12 vo2 v f = 2g ( ho − h f ) + vo2 v f = 2 ( 9.8m s 2 ) (35.0m) + (38.0 m s) 2 = 46.2 m s The Conservation of Mechanical Energy Conceptual Example: The Favorite Swimming Hole The person starts from rest, with the rope held in the horizontal position, swings downward, and then lets go of the rope. Three forces act on him: his weight, the tension in the rope, and the force of air resistance. Can the principle of conservation of energy be used to calculate his final speed? The Conservation of Mechanical Energy Example: Vertical spring with a mass A 0.20-kg ball is attached to a vertical spring. The spring constant is 28 N/m. When released from rest, how far does the ball fall before being brought to a momentary stop by the spring? The Conservation of Mechanical Energy E f = E0 1 2 mv 2f + mgh f + 12 k(Δy f )2 = 12 mv02 + mgh0 + 12 k(Δy0 )2 v0 = v f = 0 Δy0 = 0 , since spring initially unstretched Δy f = h0 hf = 0 ∴ 12 kh02 = mgh0 2mg h0 = k 2 ( 0.20 kg) ( 9.8m s2 ) = = 0.14 m 28N m Nonconservative Forces and the Work-Energy Theorem Example of a nonconservative force problem: Fireworks A 0.20 kg rocket in a fireworks display is launched from rest and follows an erratic flight path to reach the point P, as in the figure. P is 29 m above the starting point. In the process, 425 J of work is done on the rocket by the nonconservative force generated by the burning propellant. Ignoring air resistance and the mass loss due to the burning propellant, find the speed vf of the rocket at point P. THE WORK-ENERGY THEOREM Wnc = E f − E o Wnc = ( mgh f + 12 mv 2f ) − (mgh o 2 o + mv 1 2 ) Nonconservative Forces and the Work-Energy Theorem 2 f 2 o Wnc = mgh f − mgho + mv − mv 1 2 Wnc = mg ( h f − ho ) + mv 1 2 1 2 2 f 425 J = ( 0.20 kg) ( 9.80 m s2 ) ( 29.0 m ) + 12 ( 0.20 kg) v 2f v f = 61 m s Power DEFINITION OF AVERAGE POWER Average power is the rate at which work is done, and it is obtained by dividing the work by the time required to perform the work. Work W P= = Time t joule s = watt (W) Power From the work-energy theorem, doing work results in a change in energy for the system, so alternatively: Change in energy P= Time 1 horsepower = 550 foot ⋅ pounds second = 745.7 watts Another way to express the power generated by a force acting on an object moving at some average speed, P = Fv Chapter 7 Linear Momentum The Impulse-Momentum Theorem There are many situations when the force on an object is not constant. The Impulse-Momentum Theorem DEFINITION OF IMPULSE The impulse of a force is the product of the average force and the time interval during which the force acts: J = F Δt Impulse is a vector quantity and has the same direction as the average force. newton ⋅ seconds (N ⋅ s) The Impulse-Momentum Theorem J = F Δt The Impulse-Momentum Theorem DEFINITION OF LINEAR MOMENTUM The linear momentum of an object is the product of the object’s mass times its velocity: p = mv Linear momentum is a vector quantity and has the same direction as the velocity. kilogram ⋅ meter/second (kg ⋅ m/s) The Impulse-Momentum Theorem vf − vo a= Δt ∑ F = ma mv f − mv o ∑ F = Δt ! ! ! ! ! ∑ F Δt = mv f − mv o = pf − p0 ( ) The Impulse-Momentum Theorem IMPULSE-MOMENTUM THEOREM When a net force acts on an object, the impulse of this force is equal to the change in the momentum of the object impulse ! J= ! ! ! ∑ F Δt = pf − p0 ( ) final momentum initial momentum The Impulse-Momentum Theorem ! ! ! ! ∑ F Δt = pf − p0 = Δp ( ) à Divide both sides of the Impulse-Momentum Theorem by Δt ! Δp! ∴∑ F = Δt ç More general version of Newton’s 2nd Law allows for mass changing with time (e.g. rocket) ! ! ! & Δ m v !) Δp Δv ( ) = =m = ma + ( for mass constant in time → Δt Δt Δt ( + ! ! ( and then recover our familiar version + F = m a ∑ ' * The Impulse-Momentum Theorem Example: Hitting a pitched baseball. A baseball of mass 0.14 kg is pitched at a batter with an initial velocity of -38 m/s (negative is towards the bat). The bat applies an average force that is much greater than the weight of the ball, and the ball departs from the bat with a final velocity of +58 m/s. Assuming that the time of contact with the bat is 1.6 x 10-3 s, find the average force exerted on the ball by the bat. J = F Δt = mv f − mv o = (0.14)(58) - (0.14)(-38) = +13.4 kg m/s ! ! J 13.4 F= = = +8400 N −3 Δt 1.6 ×10 The Impulse-Momentum Theorem Example: A Rain Storm Rain comes down with a velocity of -15 m/s and hits the roof of a car. The mass of rain per second that strikes the roof of the car is 0.060 kg/s. Assuming that rain comes to rest upon striking the car, find the average force exerted by the rain on the roof. ∑ F Δt = mv f − mv o ( ) The Impulse-Momentum Theorem Neglecting the weight of the raindrops, the net force on a raindrop is simply the force on the raindrop due to the roof. F Δt = mv f − mv o & m # F = −$ ! v o % Δt " F = −(0.060 kg s )(− 15 m s ) = +0.90 N The Impulse-Momentum Theorem Conceptual Example: Hailstones Versus Raindrops Instead of rain, suppose hail is falling. Unlike rain, hail usually bounces off the roof of the car. If hail fell instead of rain, would the force be smaller than, equal to, or greater than that calculated in the previous Example? The Principle of Conservation of Linear Momentum WORK-ENERGY THEOREM óCONSERVATION OF ENERGY IMPULSE-MOMENTUM THEOREM ó??? Apply the impulse-momentum theorem to the midair collision between two objects….. The Principle of Conservation of Linear Momentum The midair collision between two objects. Internal forces – Forces that objects within the system exert on each other. External forces – Forces exerted on objects by agents external to the system. The Principle of Conservation of Linear Momentum ∑ F Δt = mv f − mv o ( ) OBJECT 1 ( W1 + F12 Δt = m1v f1 − m1v o1 ) External forces (gravity) ( Internal forces OBJECT 2 W2 + F21 Δt = m2 v f 2 − m2 v o2 ) The Principle of Conservation of Linear Momentum ( W1 + F12 Δt = m1v f1 − m1v o1 ) + ( ( W2 + F21 Δt = m2 v f 2 − m2 v o2 ) W1 + W2 + F12 + F21 Δt = ( m1v f1 + m2 v f 2 ) − ( m1v o1 + m2 v o2 ) ) F12 = − F21 Internal forces cancel from Newton’s 3rd law Pf Total final momentum Po Total initial momentum The Principle of Conservation of Linear Momentum The internal forces cancel out. ( W1 + W2 Δt = Pf − Po ) More generally, (sum of average external forces) Δt = Pf − Po ( ∑F EXT ) Δt = Pf − P0 The Principle of Conservation of Linear Momentum ( ∑F EXT ) Δt = Pf − P0 If the sum of the external forces is zero, then 0 = Pf − Po Pf = Po PRINCIPLE OF CONSERVATION OF LINEAR MOMENTUM The total linear momentum of an isolated system is constant (conserved). An isolated system is one for which the sum of the average external forces acting on the system is zero. The Principle of Conservation of Linear Momentum Conceptual Example: Is the Total Momentum Conserved? Imagine two balls colliding on a billiard table that is friction-free. Use the momentum conservation principle in answering the following questions. (a) Is the total momentum of the two-ball system the same before and after the collision? (b) Answer part (a) for a system that contains only one of the two colliding balls. The Principle of Conservation of Linear Momentum PRINCIPLE OF CONSERVATION OF LINEAR MOMENTUM The total linear momentum of an isolated system is constant (conserved). An isolated system is one for which the sum of the average external forces acting on the system is zero. In the top picture the net external force on the system is zero. In the bottom picture the net external force on the system is not zero. The Principle of Conservation of Linear Momentum Applying the Principle of Conservation of Linear Momentum 1. Decide which objects are included in the system. 2. Relative to the system, identify the internal and external forces. 3. Verify that the system is isolated. 4. Set the final momentum of the system equal to its initial momentum. Remember that momentum is a vector. The Principle of Conservation of Linear Momentum Example: Ice Skaters Starting from rest, two skaters push off against each other on ice where friction is negligible. One is a 54 kg woman and one is a 88 kg man. The woman moves away with a velocity of +2.5 m/s. Find the recoil velocity of the man. The Principle of Conservation of Linear Momentum Pf = Po m1v f 1 + m2 v f 2 = 0 vf 2 = − vf 2 m1v f 1 m2 ( 54 kg )(+ 2.5 m s ) =− = −1.5 m s 88 kg