Lec19

# Lec19 - Physics 2A Olga Dudko UCSD Physics Lecture 19 Today...

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Unformatted text preview: Physics 2A Olga Dudko UCSD Physics Lecture 19 Today: Physics of collisions and explosions Rotational motion Totally inelastic collision: p conserved two colliding objects stick together; conservation of momentum: m1v1 + m2v2 = (m1+m2)vf r r r m1v1i + m2v 2i Big Fish (4kg) in motion (v = 5m/s) vf = catches Little Fish (1 kg) m1 + m2 If before the collision only one 4/5 Little Fish (1kg) in motion (v = 5m/s) object (m2) is in is caught by Big Fish (4kg) motion: r m2 r v 2i vf = m1 + m2 1/5 Kinetic energy is conserved: 1 1 1 1 2 2 2 2 m1v1i + m2v 2i = m1v1 f + m2v 2 f 2 2 2 2 Conservation of momentum: m1v1i + m2v2i = m1v1f + m2v2f Head-on collision: internal forces act along the same line as the incident motion => subsequent motion is also along this line. The collision is 1D. v1i Combining the momentum (x-component) and energy conservation equations for 1D, we have: v2i Elastic collisions: K and p conserved m1 - m2 2m2 v1f = v1i + v 2i m1 + m2 m1 + m2 v2 f 2m1 m2 - m1 = v1i + v 2i m1 + m2 m1 + m2 Explosions Inverse of an inelastic collision is an Explosion where potential energy (chemical or nuclear) is released leading to increase in Kinetic energy of the fragments. Like collisions, explosions will also conserve momentum (as long as we define our system properly). Assume we have a bomb sitting on a table and then it explodes into pieces. What is the final momentum of the system? The final momentum of the system is zero. (or three). The harder cases are collisions in two dimensions Just remember that the motions in perpendicular If for external forces, Apply conservation of momentum separately to each direction. Fx = 0 and Fy = 0, then v1f sin v1f 2D Collisions directions (x and y) are independent. px and py are conserved. v1i m1 +y v1f cos +x m2 v2f sin v2f cos Before the collision After the collision v2f 2D Collisions Example A 10.0g golf ball initially traveling at 5.00m/s hits a 0.500kg stationary basketball off center. The two collide such that they finally travel at 50.0o with respect to one another, as shown in the figure. Find the final velocities of both balls. vi 30 20 Solution First, you must define a coordinate system. Let's say that the original motion of the golf ball is in the positive x-direction. Solution (cont'd) 2D Collisions Every Let's define the two balls as a single system. pix = pgx + pBx force between them is now internal. In x-direction the initial momentum of the system is: In y-direction initial momentum of the system is: piy = pgy + pBy = 0 pix = (5 m s )(0.01kg) + 0 = 0.05kg m s For the final momentum of the vg vgcos system we need to break the final velocities into components. vgsin Solution (con't) system is: 2D Collisions 30 20 In x-direction the final momentum of the p fx = pgx, f + pBx, f p fx = mg v gx + mB v Bx p fx = mg v g cos30 + mB v B cos20 Apply momentum conservation: 0.05kg m s = mgv g cos 30 + m B v B cos20 In y-direction the final momentum of the system is: p fy = pgy, f + pBy, f p fy = mg v gy + mB v By p fy = mgv g sin 30 - mB v B sin20 =0 Apply momentum conservation: piy = p fy Solution (cont'd) 2D Collisions 30 20 0 = mg v g sin 30 - mB v B sin20 mg v g sin 30 = mB v B sin20 Let's solve for vg: (0.5kg) sin20 mB sin20 vg = vB = 34.2v B vg = vB (0.01kg) sin 30 m sin30 g Substitute this into the relation for x-direction: 0.05kg m s = m gv g cos 30 + mB v B cos20 0.05kg m s = (0.01kg)( 34.2v B ) cos30 + (0.5kg)v B cos20 0.05kg m s = (0.296)v B + (0.470)v B Answer 2D Collisions 0.05kg m s = (0.766)v B 0.05kg m s = 0.0653m s vB = (0.766) Putting this back into the y-direction results: v g = 34.2v B v g = 34.2(0.0653m s ) = 2.23m s Clicker Question Impulse is: A) a force that is applied at a random time. B) a force that is applied very suddenly. C) the area under the force curve in a force-versustime graph. D) the time interval that a force lasts. E) a scalar. What's in common? xis da xe afi nd rou sa ate rot t ha yt od ab es olv inv ch Ea Rotational Variables Rotational motion is described using quantities analogous to those of linear motion. Angular displacement is the change in angular position: = f - i For ease, we will deal with an idealized model of a rigid body - one whose parts remain in fixed positions relative to one another. Each point on the rigid body undergoes the same angular displacement. Rotational Variables The convention for angular displacement is that a clockwise displacement is negative and a counterclockwise displacement is positive. Warning: can be greater than 2 rad. Angular speed occurring in the time t: avg Average angular speed is the angular displacement x = - i = t t f - ti f tf r f i When angular speed is changing, we define instantaneous angular speed: ti d = lim = t 0 t dt 0 x Units of angular speed are [rad/s]. Angular speed is positive if Angular speed is negative if is increasing (ccw). is decreasing (cw). Angular acceleration a time interval: Average angular acceleration is angular speed over avg = - i = t t f - ti f When angular acceleration is changing (as when a CD starts up after you turn on the power), we define instantaneous angular acceleration: d = lim = t 0 t dt Units of angular acceleration are [rad/s2]. Angular variables correspond to linear variables: --> x --> v --> a constant-acceleration equations apply with the proper substitutions: Equation If angular acceleration is constant, then all our Equations for constant angular acceleration Missing Quantity v = v 0 + at 1 x = (v 0 + v)t 2 1 x = v 0 t + at 2 2 = 0 + 0 t + )t 1 = ( 2 = 2 1 2 t 0t + 2 t v = v + 2a x 2 2 0 2 = v0 + 2 v equations rely on the fact that angular acceleration remains constant over the time period in question. Remember (just like in linear motion) these For Next Time: Finish up homework for Chapter 11 Study for the Quiz 6 (Ch.10&11) Read section 12-1 from Chapter 12 Start homework for Chapter 12 ...
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