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Chapter 7 Linear Momentum 老師：熊京民 教室：ME206 日期：12/4 Content 7.1 Momentum…………………………………………………3 7.2 Impulse……………………………………………………..7 7.3 Conservation of Momentum…………………………….12 7.4 Center of Mass…………………………………………...16 例題7.3、7.5、7.7……………………………………….23 7.5 Motion of the Center of Mass…………………………..29 7.6 & 7.7 Collisions in One and Two Dimension………….33 例題7.8、7.10、7.11…………………………………….41 1/46 Objectives Definition of Momentum Impulse Impulse Conservation of Momentum Center of Mass Motion of the Center of Mass Collisions (1d, 2d; elastic, inelastic) 2/46 §7.1 Momentum Consider two interacting bodies: F21 F12 Fnet If we know the net force on each body then v at t m The velocity change for each mass will be different if the masses are different. 3/46 Rewrite the previous result for each body as: m1v1 F21t m2 v 2 F12 t F21t Combine the two results: m1v1 m2 v 2 m1 v1 f v1i m2 v 2 f v 2i 4/46 The quantity (mv) is called momentum (p). p p = mv and is a vector v vector. The unit of momentum is kg m/s; there is no derived unit for momentum. 5/46 From slide (3): m1v1 m2 v 2 p1 p 2 The change in momentum of the two bodies is “equal and equal opposite”. opposite Total momentum is conserved during the interaction; the momentum lost by one body is gained by the other. 6/46 §7.2 Impulse Definition of impulse: p Ft One can also define an average impulse when the force is variable. 7/46 Example (text conceptual question 7.2): A force of 30 N is applied for 5 sec to each of two bodies of different masses. 30 N Take m1<m2 m1 or m2 (a) Which mass has the greater momentum change? Since the same force is applied to p Ft each mass for the same interval, p is the same for both masses. 8/46 Example continued: (b) Which mass has the greatest velocity change? p Since both masses have the same p, v the smaller mass (mass 1) will have the m larger change in velocity. (c) Which mass has the greatest acceleration? v Since av the mass with the greater a velocity change will have the greatest t acceleration. 9/46 Example (text problem 7.9): An object of mass 3.0 kg is Example allowed to fall from rest under the force of gravity for 3.4 seconds. What is the change in momentum? Ignore air resistance. Want p =mv. v at v gt 33.3 m/sec p mv 100 kg m/s (downward) 10/46 Example (text problem 7.10): What average force is necessary to bring a 50.0-kg sled from rest to 3.0 m/s in a period of 20.0 seconds? Assume frictionless ice. p Fav t p mv Fav t t Fav 50.0 kg3.0 m/s 7.5 N The force will be 20.0 s in the direction of motion. 11/46 §7.3 Conservation of Momentum v1i v2i m1>m2 m1 m2 A short time later the masses collide. m1 m2 What happens? 12/46 During the interaction: y N1 N2 x F21 F12 w1 w2 F y N1 w1 0 F y N 2 w2 0 F x F21 m1a1 F x F12 m2 a2 There is no net external force on either mass. 13/46 The forces F12 and F21 are internal forces. This means that: p1 p 2 p1 f p1i p 2 f p 2i p1i p 2i p1 f p 2 f In other words, pi = pf. That is, momentum is conserved. This statement is valid during the interaction only. 14/46 Example (text problem 7.18): A rifle has a mass of 4.5 kg and it fires a bullet of 10.0 grams at a muzzle speed of 820 m/s. What is the recoil speed of the rifle as the bullet leaves the barrel? As long as the rifle is horizontal, there will be no net external force acting on the rifle-bullet system and momentum will be conserved. pi p f 0 mb vb mr vr mb 0.01 kg vr 4.5 kg 820 m/s 1.82 m/s vb mr 15/46 §7.4 Center of Mass The center of mass (CM) is the point representing the mean CM (average) position of the matter in a body. This point need average not be located within the body. 16/46 The center of mass (of a two body system) is found from: system m1 x1 m2 x2 xcm m1 m2 This is a “weighted” average of the positions of the particles that compose a body. (A larger mass is more important.) 17/46 m r i i where mtotal mi i In 3-dimensions write rcm m i i i The components of rcm are: m x i i m y i i m z i i i ycm i zcm i m xcm m i m i i i i i 18/46 Example (text problem 7.27): Particle A is at the origin and has a mass of 30.0 grams. Particle B has a mass of 10.0 grams. Where must particle B be located so that the center of mass (marked with a red x) is located at the point (2.0 cm, 5.0 cm)? y x ma xa mb xb mb xb xcm ma mb ma mb ma ya mb yb mb yb ycm A x ma mb ma mb 19/46 Example continued: 10 g xb ycm 10 g yb 5 cm xcm 2 cm 10 g 30g 30 g 10 g xb 8 cm yb 20 cm 20/46 Example (text problem 7.30): The positions of three particles are (4.0 m, 0.0 m), (2.0 m, 4.0 m), and (-1.0 m, -2.0 m). The masses are 4.0 kg, 6.0 kg, and 3.0 kg respectively. What is the location of the center of mass? y 2 1 x 3 21/46 Example continued: m1 x1 m2 x2 m3 x3 xcm m1 m2 m3 4 kg4 m 6 kg2 m 3 kg 1 m 4 6 3 kg 1.92 m m1 y1 m2 y2 m3 y3 ycm m1 m2 m3 4 kg0 m 6 kg4 m 3 kg 2 m 4 6 3 kg 1.38 m 22/46 例題 7.3 7.5 7.7 23/46 24/46 25/46 26/46 27/46 Example 7.07b 28/46 §7.5 Motion of the Center of Mass For an extended body, it can be shown that p = mvcm. From this it follows that Fext = macm. 29/46 Example (text problem 7.35): Body A has a mass of 3.0 kg and vx = +14.0 m/s. Body B has a mass of 4.0 kg and has vy = -7.0 m/s. What is the velocity of the center of mass of the two bodies? Consider a body made up of many different masses each with a mass mi. The position of each mass is ri and the displacement of each mass is ri = vit. 30/46 Example continued: m r i i For the center of mass: rcm v cm t i m i i Solving for the velocity of the center of mass: ri mi t m v i i v cm i i mi i m i i Or in component form: m v i i,x m v i i, y v cm,x i v cm,y i m i i m i i 31/46 Example continued: Applying the previous formulas to the example, ma va , x mb vb , x vcm, x ma mb 3 kg 14 m/s 4 kg 0 m/s 6 m/s 3 4 kg ma va , y mb vb , y vcm, y ma mb 3 kg 0 m/s 4 kg 7 m/s 4 m/s 3 4 kg 32/46 §§ 7.6 & 7.7 Collisions in One and Two Dimension When there are no external forces present, the momentum of a system will remain unchanged. (pi = pf) If the kinetic energy before and after an interaction is the same, the “collision” is said to be perfectly elastic If the elastic. inelastic. kinetic energy changes, the collision is inelastic 33/46 together, If, after a collision, the bodies remain stuck together the maximum. loss of kinetic energy is a maximum This type of collision is called perfectly inelastic. inelastic 34/46 Example (text problem 7.41): In a railroad freight yard, an empty freight car of mass m rolls along a straight level track at 1.0 m/s and collides with an initially stationary, fully loaded, boxcar of mass 4.0m. The two cars couple together upon collision. (a) What is the speed of the two cars after the collision? pi p f p1i p2i p1 f p2 f m1v1 0 m1v m2 v m1 m2 v m1 m m v1 0.2 m/s v 1 2 35/46 Example continued: (b) Suppose instead that both cars are at rest after the collision. With what speed was the loaded boxcar moving before the collision if the empty one had v1i = 1.0 m/s. pi p f p1i p2i p1 f p2 f m1v1i m2 v2i 0 0 m1 v2i v1i 0.25 m/s m 2 36/46 Example (text problem 7.49): A projectile of 1.0 kg mass approaches a stationary body of 5.0 kg mass at 10.0 m/s and, after colliding, rebounds in the reverse direction along the same line with a speed of 5.0 m/s. What is the speed of the 5.0 kg mass after the collision? pi p f p1i p2i p1 f p2 f m1v1i 0 m1v1 f m2 v2 f v2 f m1 v1i v1 f m2 1.0 kg 10 m/s 5.0 m/s 3.0 m/s 5.0 kg 37/46 Example (text problem 7.58): Body A of mass M has an original velocity of 6.0 m/s in the +x-direction toward a stationary body (B) of the same mass. After the collision, body A has vx=+1.0 m/s and vy=+2.0 m/s. What is the magnitude of body B’s velocity after the collision? Final Initial A A vAi B B 38/46 Example continued: x momentum: y momentum: pix p fx piy p fy p1ix p2ix p1 fx p2 fx p1iy p2iy p1 fy p2 fy m1v1ix 0 m1v1 fx m2 v2 fx 0 0 m1v1 fy m2 v2 fy Solve for v2fx: Solve for v2fy: m1v1ix m1v1 fx m1v1 fy v2 fx v2 fy v1 fy m2 m2 v1ix v1 fx 2.00 m/s 5.00 m/s The mag. of v2 is v2 f v 2 2 fy v 2 2 fx 5.40 m/s 39/46 Summary •Definition of Momentum •Impulse •Center of Mass •Conservation of Momentum 40/46 例題 7.8 7.10 7.11 41/46 42/46 Example 7.08b 43/46 Example 7.10 44/46 Example 7.11a 45/46 Example 7.11b 46/46 第七章完