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Collisions and rotational kinematics • Spring break starts tomorrow! • Finish up collisions today • Start Chapter 10 (Rotational Motion). 1 Collision summary ! ! ! ! In all collisions, mA vAi + mBvBi = mA vAf + mBvBf momentum is conserved: This is a vector equation. Components are conserved. mA vAxi + mBvBxi = mA vAxf + mBvBxf mA vAyi + mBvByi = mA vAyf + mBvByf For elastic collisions, kinetic 2 1 energy is conserved. mA vAi + 2 In perfectly inelastic collisions, ! !final velocities are the same vAf = vBf 1 2 2 2 mBvBi2 = 12 mA vAf + 12 mBvBf vAyf = vByf Components: vAxf = vBxf Remember momentum and velocities have signs However kinetic energy is always positive 2 Elastic collision example Two air carts have equal mass and springs attached. Cart A is moving toward cart B with speed vAi. Cart B is initially at rest. What is the speed of the carts after the collision? The carts are on a track and have springs (conservative force). This is an elastic collision in 1D which gives us two equations: mAvAi + mBvBi = mAvAf + mBvBf 1 1 1 1 2 2 2 2 m v + m v = m v + m v 2 A Ai 2 B Bi 2 A Af 2 B Bf Since masses are equal, can cancel them out. Also, vBi=0 2 2 2 vAi = vAf + vBf vAi = vAf + vBf 2 2 2 2 + 2vAf vBf + vBf = vAf + vBf or 2vAf vBf = 0 Substituting gives vAf Either vAf or vBf is 0 and the other must equal vAi. Since the carts can’t pass through each other, cart A must stop and 3 cart B travels with velocity vAi. Clicker question 1 Set frequency to BA m A big ball, mass M=10m, speed v, v 10m strikes a small ball, mass m, at rest. Could the following occur: The big ball comes to a complete stop and the 10m small ball takes off with speed 10v? A. Yes, this could occur B. No, it would violate conservation of momentum C. No, it would violate conservation of energy D. Need more information. Momentum Pi = 10mv is conserved: Pf = 10mv m 10v 2 2 1 K = 10mv = 5mv Kinetic energy is i 2 not conserved: K f = 1 m(10v)2 = 50mv 2 2 We can lose kinetic energy during a collision (that is the definition of an inelastic collision) but we cannot gain kinetic energy. 4 Other examples of elastic collisions Newton’s cradle is a good example of elastic collisions Need to conserve both kinetic energy and momentum The basketball and tennis ball experiment is a little complicated to work out so I will just set it up and then jump to the answer. The full details will be in the posted slides. The text book works out the results of elastic collisions in more depth. 5 An elastic example A tennis ball is put on top of a basketball and both are dropped from a height of 1 m. How high will the tennis ball go? Assume all collisions are elastic. Can use conservation of energy to get velocity after falling 1 m. K1 +U1 = K 2 +U 2 becomes mgh = 12 mv 2 so v = 2gh = 2 ⋅10 m/s2 ⋅1 m = 4.5 m/s Basketball hits first. What happens? If no friction losses then by conservation of energy the upward speed after bounce equals the downward speed before bounce Or, elastic collisions conserve kinetic energy. Same answer. Is momentum conserved? Only if you consider Earth as part of the system An elastic example So we have a 1D elastic collision between basketball moving at vB = 4.5 m/s and tennis ball moving at vT = –4.5 m/s The collision conserves momentum: mBvB1 + mT vT1 = mBvB2 + mT vT2 Elastic collision conserves kinetic energy: 1 2 2 2 2 2 mBvB1 + 12 mT vT1 = 12 mBvB2 + 12 mT vT2 Solve this problem in the reference frame where the basketball is at rest This gives mT vT1 = mBvB2 + mT vT2 and vB1 = 0 m/s vT1 = −9.0 m/s 2 2 2 mT vT1 = mBvB2 + mT vT2 An elastic example 2 2 2 m v = m v + m v m v = m v + m v We have T T1 B B2 T T2 and T T1 B B2 T T2 2 2 2 2 2 mBvB2 = mT vT1 − mT vT2 = mT ( vT1 − vT2 ) = mT (vT1 − vT2 ) (vT1 + vT2 ) Factoring: mBvB2 = mT vT1 − mT vT2 = mT ( vT1 − vT2 ) Divide the equations to get vB2 = vT1 + vT2 Substitute in to get mBvT1 + mBvT2 = mT vT1 − mT vT2 Rearrange: mT vT2 + mBvT2 = mT vT1 − mBvT1 becomes ( mT + mB ) vT2 = ( mT − mB ) vT1 ( mT − mB ) vT1 Solve: vT2 = (mT + mB ) Substitute back in to get ( mT − mB ) 2mT vB2 = vT1 + vT2 = vT1 + (mT + mB ) vT1 = mT + mB vT1 An elastic example mT − mB ) So in the basketball rest frame we have ( vT2 = vT1 a post collision tennis ball velocity of ( mT + mB ) Masses and initial velocities: mB = .4 kg vB1 = 0 m/s mT = .07 kg vT1 = −9.0 m/s .07 kg −.4 kg) ( vT2 = (−9.0 m/s) = 6.3 m/s (.07 kg + .4 kg) This is with respect to the basketball which is moving up at 4.5 m/s. Therefore the tennis ball speed with respect to the ground is 6.3 m/s + 4.5 m/s = 10.8 m/s. Basketball and tennis ball Before collision, the tennis ball is moving down at 4.5 m/s After collision the tennis ball is moving up at 10.8 m/s What height will the tennis ball reach? 1 mv2 = mgh so h = v 2 2g 2 2 ( 10.8 m/s) = = 5.8 m 2 ⋅10 m/s2 19 foot height from just a 3 foot drop?! Inelastic vs elastic collisions We have two balls with different material properties: The sad ball just goes squash and lies there like a lump The happy ball bounces right back like a champ We will use each ball connected to the end of a pendulum to try to knock over a block of wood. Clicker question 2 Set frequency to BA Two balls of equal mass m are used as the bob of a pendulum and are released from the same point. Which ball is more likely to knock over the block of wood? A. The ball that splats on the wood (sad ball) B. The ball that bounces off the wood (happy ball) C. It doesn’t make a difference which ball is used Both balls have the same mass and velocity just before hitting the block and therefore the same momentum: p = mv Conservation of momentum: mballvball,i + 0 = mballvball,f + mblock vblock,f Sad ball ends with p ≈ 0 so by conservation of momentum, the block gains momentum Δp ≈ mv (it gets an impulse of mv). Happy ball ends with p ≈ −mv (because it bounces off). By conservation of momentum, the block gains momentum of Δp ≈ 2mv. Angular kinematics Early in the semester, we dealt with kinematics which involved displacement, velocity, and acceleration. Angular kinematics is the same thing but for objects which are rotating (rather than translating). For something to rotate, it must have an axis about which it rotates like the axle for a wheel. Only sensible place for the origin is along the axis. Use polar coordinates (r,θ) instead of Cartesian coordinates (x,y). Axis is in the z-direction. r θ 13 Angular kinematics θ is a measure of angular position Δθ is a measure of angular displacement Δθ r θ In angular motion (and many other areas of physics) it is much more convenient to use radians (abbreviated rad) rather than degrees or revolutions to measure angles. To convert you just need to remember 1 revolution (1 rev) = 2π radians (2π rad) = 360 degrees (360°) 14 Angular velocity Angular velocity tells us how fast (and in what direction) something is spinning. Δθ ω z, avg = Δt ωz = Δθ r θ dθ dt The z subscript indicates the axis is in the z-direction (and the rotation is therefore in the xy plane) Counterclockwise is positive Also have angular acceleration which describes how the spinning rate changes Δω dω d 2θ α z, avg = Δt αz = dt = dt 2 15 Angular kinematics The same equations which were derived for constant linear (also known as translational) acceleration apply for constant angular (also known as rotational) acceleration Constant linear acceleration only! vx = v0 x + at x = x0 + v0 x t + 12 ax t 2 2 x 2 0x v = v + 2ax ( x − x0 ) Constant angular acceleration only! ω z = ω0 z + α zt θ = θ 0 + ω 0 z t + 12 α z t 2 ω z2 = ω 02z + 2α z (θ − θ 0 ) 16 Relationships to linear velocity If we want the linear displacement or velocity of a point on a rotating object, we need r and either θ or ω. What is the speed of the tire rim if the radius r s is 0.35 m and the tire rotates at 20 rad/s? θ Radians measure distance around a unit circle Therefore s = rθ (only for θ in radians!) ds dθ =r Time derivative of both sides (r is constant): dt dt v = rω . Linear speed is radius times angular speed Rim speed is v = rω = 0.35 m ⋅ 20 rad/s = 7 m/s 17