ics Tuesday, ober 05, 2004 Ch 6: Ch 9: Last Example Impulse Momentum
Announcements Help this week: Wednesday, 8-9 pm in NSC 118/119 Sunday, 6:30-8 pm in CCLIR 468
Announcements This week s lab will be a workshop. Bring your Ranking Tasks workbook and your textbook.
Ch 6: Applying Newton s Laws Note the following: How does your weight change while you re Ascending at v = const.? Descending at v = const.? Accelerating Down? Accelerating Up?
Ch 6: Applying Newton s Laws Two blocks of mass 3.50 kg and 8.00 kg are connected by a massless string passing over a frictionless pulley. The inclines are also frictionless. Find (a) the magnitude of the acceleration of each block and (b) the tension in the string. m 1 =3.50 kg m 2 = 8.00 kg Problem Sheet #1 θ 1 =35 o θ 2 =35 o? (P1) Inclined plane & newton s
Ch 6: Applying Newton s Laws Two blocks of mass 3.50 kg and 8.00 kg are connected by a massless string passing over a frictionless pulley. The inclines are also frictionless. Find (a) the magnitude of the acceleration of each block and (b) the tension in the string. FBDs For m 1 For m 2 n 1 T T n 2 W 1 W 2
Ch 6: Applying Newton s Laws Pictorial Representation: y 1 x2 y θ 1 =35 o θ 2 =35 o 2 Unknowns: T, a 1,x1 x 1 Knowns: Δx 1 = Δx 2 v 1,x1 = v 2,x2 a 1,x1 = a 2,x2 m 1 = 8.00 kg m 2 = 3.50 kg θ 1 = 35 0 θ 2 = 35 0 g = 9.81 m/s 2 W 1x = m 1 g (sinθ 1 ) W 1y = - m 1 g (cosθ 1 ) W 2x = - m 2 g (sinθ 2 ) W 2y = - m 2 g (cosθ 2 ) a 1,y1 = a 2,y2 = 0
Ch 6: Applying Newton s Laws Mathematical Representation: Let s look at Newton s 2 nd Law in the x-direction for both blocks. a 1,x1 = F x1,net m 1 = m 1 g sinθ 1 T m 1 a 2,x 2 = F x2,net m 2 = T m 2 g sinθ 2 m 2 These accelerations must be the same!
Ch 6: Applying Newton s Laws Mathematical Representation: m 1 g sinθ 1 T m 1 = T m 2 g sinθ 2 m 2 Rearranging and using the fact that θ 1 = θ 2. T 1 + 1 m m 1 2 2g sinθ = 0
Ch 6: Applying Newton s Laws Finally, plugging in numbers. T 1 3.50kg + 1 8.00kg 2(9.81 m )sin35 0 = 0 s 2 T = 27.4 N Now plugging back into Newton s 2nd Law we find a 1,x1 = (3.50kg)(9.81m s 2 )sin35 0 27.4N (3.50kg) = 2.20 m s 2
Ch 9: Momentum and Collisions Linear Momentum & Collisions In the study of systems with multiple objects, particularly systems in which these objects interact with one another through collisions, it is useful to take advantage of the concept of momentum. We ll see that this new quantity is conserved in isolated systems, allowing us to analyze some rather complicated interactions.
Ch 9: Momentum & Collisions We ve applied Newton s Laws to better understand the behavior of systems - single object - multiple object - remote interactions Now, let s examine interactions between objects through another approach:
Ch 9: Momentum & Collisions What constitutes an impulse is probably best illustrated through a simple demonstration. When I drop the ball, the force of gravity exerts an impulse on the falling ball. When the ball hits the floor, a second force also exerts an impulse: the contact force of the floor.
Ch 9: Momentum and Collisions For a constant force, as with the falling ball, it s easy! I = FΔt Force (N) t i Time (s) t f Notice graphically that the impulse is the area under the curve of a force vs time graph. (Just like distance was the area under a velocity vs time graph!) Graphical interpretation of
Ch 9: Momentum and Collisions I = FΔt [ I ] = [ F][Δt] [ I ] = N s = kg m s
Ch 9: Momentum and Collisions So, what about non-constant forces? Force (N) Average Force Time (s) I = ( F f + F i ) 2 Δt = Let s look at a simple case of a force that increases linearly with time. F Δt The answer is still just the area under the curve.
Ch 9: Momentum and Collisions So, what about non-constant forces? Force (N) I floor = ( F max + F min ) 2 Average Force Time (s) Δt = F floor Δt Let s assume that during the bounce, the floor exerts a force on the ball as given at left. Note: The force of gravity results in an impulse at the same time.
Ch 9: Momentum & Collisions A 0.200-kg rubber ball is released from a height of 2.00 m above the floor. The ball falls, bounces, and rebounds. How high does the ball make it on the rebound if the force function is given by the graph below? 480 N Problem Sheet #2 F floor (N) 5 ms 10 ms 15 ms Time (s)
Ch 9: Momentum & Collisions This problem needs to be separated into 3 parts: 1) the ball falling from 2.00 m to the floor 2) the interaction of the ball with the floor 3) the ball rising to a new, unknown height Motion Diagrams: (1) (2) Close-Up (3) v a v a a v? (P2) bouncing bal - Motion
Ch 9: Momentum & Collisions Pictorial Representation (1): Knowns: y i = 2.00 m, y f = 0 v i = 0 t i = 0 a = - g m = 0.200 kg y i, v yi, t i a +y Unknowns: t f, v f y f, v yf, t f
Ch 9: Momentum & Collisions Mathematical Representation (1): Knowns: y i = 2.00 m, y f = 0 v i = 0 t i = 0 a = - g m = 0.200 kg 0 = 2.00m + 0 + 1 2 ( 9.81m s 2 )(Δt) 2 Δt = t f - t i = t f - 0 = t f = 0.639 s Unknowns: t f, v f v y, f = v i,y + a y (Δt) v y, f = 0 + ( 9.81 m s 2 )(0.639s) = 6.26 m s
Ch 9: Momentum & Collisions Free-Body Diagram (2): From the graph of force vs time and our definition of impulse, we can find an average force over the 10 ms of the bounce. We ll base our FBD on this average force. n floor W Mathematical Representation (2): Newton s 2 nd law a y = F y,net m = n W m = n mg m
Ch 9: Momentum & Collisions Mathematical Representation (2): To get the normal force, we need to use the graph and our definition of impulse. I = area under curve = (F + F ) max min (Δt) 2 (480N 0) I = (15ms 5ms) = 2.4Ns 2 480 N F floor (N) 5 ms 10 ms 15 ms Time (s)
Ch 9: Momentum & Collisions Mathematical Representation (2): To get the normal force, we need to use the graph and our definition of impulse. F floor = I floor (Δt) = +2.4Ns 10ms = 240N 480 N F floor (N) 5 ms 10 ms 15 ms Time (s)
Ch 9: Momentum & Collisions Mathematical Representation (2): n floor Back to Newton s 2 nd law a y = F y,net m = n W m = n mg m a y = (240N) (0.200kg)(9.81m s 2 ) 0.200kg W = 1190 m s 2
Ch 9: Momentum & Collisions Pictorial Representation (2): Knowns: y i = 0 m, y f = 0 v i = - 6.26 m/s t i = 0.639 s t f = 0.649 s a y = +1190 m/s 2 m = 0.200 kg Unknowns: v f y i, v yi, t i y f, v yf, t f Note: The initial state of this part (2) of the problem was the final state of the previous part (1). a +y
Ch 9: Momentum & Collisions Mathematical Representation (2): Knowns: y i = 0 m, y f = 0 v i = - 6.26 m/s t i = 0.639 s t f = 0.649 s a y = +1190 m/s 2 m = 0.200 kg Unknowns: v f v y, f = v i,y + a y (Δt) v y, f = 6.26 m s + (1160 m s 2 )(0.649s - 0.639s) v y, f = 5.64 m s
Ch 9: Momentum & Collisions Pictorial Representation (3): Knowns: y i = 2.00 m v i = +5.64 m/s v f = 0 t i = 0.649 s a = - g m = 0.200 kg y f, v yf, t f y i, v yi, t i a +y Unknowns: t f, y f
Ch 9: Momentum & Collisions Mathematical Representation (3): Knowns: y i = 2.00 m v i = +5.64 m/s v f = 0 t i = 0.649 s a = - g m = 0.200 kg v y, f = v i,y + a y (Δt) 0 = (5.64 m s) + ( 9.81 m s 2 )(t f 0.649s) t f = 1.224 s Unknowns: t f, y f y f = 0 + (5.64 m s)(1.224s 0.649s) + 1 2 ( 9.81m s 2 )(1.224s 0.649s) 2 y f = 1.62 m
Ch 9: Momentum and Collisions So, what about non-constant forces? We can always find the a constant F such that the impulse delivered is the same as that delivered by a non-constant force. Force (N) Force (N) Time (s) Time (s)
Ch 9: Momentum and Collisions So, what about non-constant forces? F net Force (N) = area under curve Δt Force (N) The constant value is always exactly the timeaveraged value of the nonconstant force! Time (s) Time (s)
Ch 9: Momentum and Collisions Completely General Having explored this quantity graphically and using results from calculus (not shown), we can say succinctly and generally that: I = F Δt net
Ch 9: Momentum and Collisions The linear momentum of an object of mass m moving with velocity v is defined as the product of the mass and the velocity: Notice that momentum is a vector quantity, which means that it must be specified with both a magnitude and direction. p = mv Also notice that the direction of the momentum is necessarily parallel to the velocity.
Ch 9: Momentum and Collisions p = mv [ p] = [m][ v] [ p] = kg (m/s) OR [ p] = N s The units suggest a relationship between force and momentum.
Ch 9: Momentum and Collisions p = mv It accelerates. What happens when we apply a force to an object? Its velocity changes. Its momentum changes. The force imparts momentum.
Ch 9: Momentum and Collisions p = mv In fact, we can relate force to momentum changes directly using Newton s 2nd Law: F = ma = m Δ v net Δt F = Δ p net Δt This form is completely general, as it allows for changes in mass as well as velocity with time.