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Conservation of Momentum Conservation of momentum: m1v1i m2v2i m1v1 f m2v2 f Split into components: m1v1ix m2v2ix m1v1 fx m2v2 fx m1v1iy m2v2iy m1v1 fy m2v2 fy If the collision is elastic, we can also use conservation of energy. Two Methods Method 1 - Graphical Method – Ruler and protractor. 1. Calculate the momentum of each object. 2. Choose an appropriate scale. Draw each for the momenta before the interaction Arrange them head to tail Draw the resultant, the total momentum before 3. Draw each for the momenta after the interaction Arrange them head to tail Draw the resultant, the total momentum after 4. Compare the magnitude and direction of the total momentum vectors before and after. If they are identical momentum is conserved. Method 2: Trigonometric Method 1. 2. 3. 4. 5. Calculate the momentum of each from p = mv. Use trig calculate the X and Y components of each object before. Find the x and y components after the collision. Total the X components before the collision opp sin Total the X components after the collision hyp Set them equal to each other Total the Y components before the collision adj cos Total the Y components after the collision hyp Set them equal to each other in magnitude and direction Solve the individual component equations for the unknowns. opp tan Make a chart and fill out as in the following example ady 1. Fill out the chart pxi pyi pxf pyf Mass 1 Mass 2 Total p p yi xi p xf p yi 2. Solve the component equations pxi pxf p yi p yf Two types of 2-D collisions Type 1 – Glancing collision - One object hits another stationary object they fly off in separate directions. Ex: The masses of the green ball and red ball are 2 kg and 3 kg. If the initial velocity of the green ball is 10m/s. Construct the component equations, don’t solve them. pxi pyi pxf pyf Mass 1 Mass 2 Total pxi pxf p yi p yf After the collision Before the collision for green mass for red mass for green mass p xi mvxi pxf mvrf cos 300 pxf mvgf cos 600 (2 kg )(10 m / s ) m 20kg s p yi 0 For red mass p xi 0 p yi 0 (3kg )(v f ) cos 600 (2kg )(vgf ) cos 600 1vrf 1vgf p yf mv f sin 300 p yf mvgf sin 600 (3kg )vr f sin 600 (2kg )(vgf ) sin 600 1.73 vrf 1.73vgf Type 2 – Right angle collisions Type 2 – Right angle collisions 1. Objects approach at 900, one with momentum in the x-direction the other in the y-direction only. 2. They fly off together at an angle , with some final velocity Right angle collisions A 95 kg Physics student running south the hall with a speed of 3m/s tackles a 90 kg Chemistry student moving west with a speed of 5 m/s. If the collision is perfectly inelastic, calculate: 1. The velocity of the students just after the collision 2. The energy lost as a result of the collision. Use Pythagorean Theorem and trig N pti pxi 2 p yi 2 (270kgm / s) 2 (285kgm / s) 2 m1=95kg 72,900 81, 225 V1i=3m/s 393 kgm / s V2i=5m/s W m2=90kg E p yi 285 tan 46.50 pxi 270 S Right angle collisions An eastern kingbird and a bee approach each other at right angles. The bird has a mass of .13 kg and a speed of .6 m/s, and the bee has a mass of 5x10-3 kg and a speed of 15 m/s. If the bird catches the bee, what’s the new speed of the bird? Accident investigation. Two cars of equal mass approach an intersection. One is traveling east at 29 mi/h (13.0 m/s) the other north with unknown speed. The vehicles collide and stick together, leaving skid marks at an angle of 55º north of east. The second driver claims he was driving below the speed limit of 35 mi/h (15.6 m/s). 1. Is he telling the truth? 2. What is the speed of the“combined vehicles” after the collision? 3. How long are the skid marks (uk = 0.5) Right angle collisions 13.0 m/s ?? m/s Glancing Collision A cue ball traveling at 4m/s makes a glancing, elastic collision with a target ball of equal mass initially at rest. It is found that the cue ball is deflected such that it makes an angle of 30 with respect to its original direction of travel. Set up the conservation of momentum equations in the x and y directions, don’t solve them. Glancing Collision A 2 kg ball is initially at rest on a horizontal frictionless surface. A 4 kg ball moving horizontally with a velocity of 10 m/s hits it. After the collision the 4 kg ball has a velocity of 6 m/s at an angle of 37 to the positive x-axis. 1. Determine the speed and direction of the 2 kg ball after the collision. 2. How much K.E. was lost in the collision? KE1 + KE2 = KE1 + KE2 + p2y p1y + p 2y = p1y 1 2 1 1 mv1 mv12 mv22 2 2 2 + p2x p1x + p2x =p1x 0 = m1v1sinθ1 + m2 v2sinθ2 m1v1 = m1v1cosθ1 + m2 v2cosθ2 Two skaters collide and embrace, in a completely inelastic collision. Thus, they stick together after impact, where the origin is placed at the point of collision. Alfred, whose mass mA is 83 kg, is originally moving east with speed vA = 6.2 km/h. Barbara, whose mass mB is 55 kg, is originally moving north with speed vB = 7.8 km/h.