Collision Theory 8.01t Nov 1, 2004Momentum and Impulse • momentum p G = mv G G m• change in momentum ∆p =∆vG GG • average impulse I = F ∆ =∆p G tave aveConservation of Momentum • The total change in momentum of a system and its surroundings between the final state and the initial state is zero, G G∆psystem +∆psurroundings = 0Conservation of Momentum • completely isolate system from the surroundings total = ext G F G0 • change in momentum of the system is also zero G∆p = system G0Conservation of Momentum: Isolated System When the total external force on a system is zero, then the total initial momentum of the system equals the total final momentum of the system, pGtotal Gtotal 0 = p fProblem Solving Strategies: Momentum Diagram • Identify the objects that compose the system • Identify your initial and final states of the system • Choose symbols to identify each mass and velocity in the system. • Identify a set of positive directions and unit vectors for each state. • Decide whether you are using components or magnitudes for your velocity symbols.Momentum Diagram Since momentum is a vector quantity, identify the initial and final vector components of the total momentum Initial State Gp total 0 1 2 G+m v m v= 1,0 G 2,0 + ... total x-comp: p = m1 ( v + ...x,0 x )1,0 + m2 ( vx )2,0 total y-comp: py,0 = m1 ( vy )1,0 + m2 ( vy )+ ... 2,0 Final State Gtotal p f 11, f + m2 G m v= G v 2, f + ... totalx-comp: p , = m1 ( v + ...xf x )1, f + m2 ( vx )2, f total + ... ,y-comp: pyf = m1 ( vy )1, f + m2 ( vy )2, fStrategies: Conservation of Momentum • If system is isolated, write down the condition that momentum is constant in each direction total totalp x f x,0 = p , mv1 ()+m2 (v +=mv + m2 (v +...xx 1,0 x )2,0 ... 1 (x )1, f )2, f total totalp y f y,0 = p , mv+m2 (vy )2,0 +=mvy )+...y1 ()1,0 ... 1 ( 1, f + m2 (vy )2, fPlanar Collision Theory: Energy Types of Collisions in Two Dimensions: • Elastic: Kf = Ki • Inelastic: Kf < Ki • Completely Inelastic: Only one body emerges • Superelastic: Kf > KiElastic Collisions • Kinetic Energy does not change. K0 total = Kftotal 1 2 + 1 21 m v mv2,0 2 + ... = 1 m v + mv2, f 2 + ... 2 1 1,0 2 1 1, f 2 2 2 2PRS Question Suppose a golf ball is hurled at a heavy bowling ball initially atrest and bounces elastically from the bowling ball. After the collision, 1. The golf ball has the greater momentum and the greater kinetic energy. 2. The bowling ball has the greater momentum and the greaterkinetic energy. 3. The golf ball has the greater momentum but has the smaller kinetic energy. 4. The bowling ball has the greater momentum but has the smaller kinetic energy.PRS Question Consider the exothermic reaction (final kinetic energy is greater than the initial kinetic energy). H + H → H2 + 5ev Two hydrogen atoms collide and produce a diatomic hydrogen molecule. Using only the principles of classical mechanics, this reaction 1. Is possible.2. either violates conservation of energy or conservation of momentum but not both 3. satisfies conservation of energy and momentum but is not possible for other reasons.Worked Example: Two Dimensional Elastic Collision Consider an elastic collision between two particles. In the laboratory reference frame, the first particle with mass m1, the incident particle, is moving with an initial given velocity v10. The second ‘target’ particle is of mass m2 = m1 and at rest. After the collision, the first particle movesoff at a measured angle θ1f = 45o with respect tothe initial direction of motion of the incident particle with an unknown final velocity v1f. Particle two moves off at unknown angle θ2f with an unknown final velocity v2f. Find v1f ,v2f , and θ2f.Momentum Diagram • Momentum is a vector!Analysis Momentum is constant: x-direction ˆi : mv 1 2, f cos θ2, f + mv11,0 = mv 1 1, f cos θ1, f y-direction ˆj:0 = mv sin θ2, f − mv1 2, f 1 1, f sin θ1, f Elastic collision: 1 mv 2 = 1 mv 2 1 1, f2 11,0 2 1 2, f 2 + 1 mv 2Strategy Three unknowns: θ2f , v1f, and v2f 1. Eliminate θ2f by squaring momentum equations andadding equations and solve for v2f in terms of v1f 2. Substitute expression for v2f kinetic energy equationand solve possible quadratic equation for v1f 3. Use result to find expression for v2f 4. Divide momentum equations to obtain expression for θ2fCarry out Plan: Step 1 • x-momentum: v2, f cos θ2, f = v1,0 − v1, f cos θ1, f v2 2 2 2 2, f cos θ2, f = v1, f cos θ1, f − 2v1, f cos θ1, fv1,0 + v1,0 • y-momentum: v2, f sin θ2, f = v1, f sin θ1, f 2 2 2 2v2, f sin θ2, f = v1, f sin θ1, f 2 2• Add using sin θ+ cos θ= 1 v2 2 2 2, f = v1, f − 2v1, f cos θ1, fv1,0 + v1,0 2Carry out Plan: Step 2 1 mv 2 = 1 mv 2 1 1, f• Energy: 2 11,0 2 1 2, f 2 + 1 mv 2 • Substitute result from momentum: 2 2 2 02vv1,0 = v2, f + v1, f 2 = 2v1, f − 2v1, f cos θ1, fv1,0 + v1,0 1, f 2 − 2v1, f cos θ v1,0 = 1, f • Easy to solve (no quadratic) for v1f: v1, f = v1,0 cos θ1, f 2Carry out Plan: Step 3• Use results from step 1 to solve for v2f:()12222, 1, 1, 1, 1,0 1,02cosff f fvv v vvθ=− +()()()12222, 1,0 1, 1,0 1, 1, 1,0 1,0cos 2 cos cosff ffvv v vvθθθ=− +()1222, 1,0 1, 1,0 1,1 cos sinfffvv vθθ=− =Carry out Plan: Step 4• Divide momentum equations to solve for θ2f:• Substitute result for v1f :2, 2, 1,0 1, 1,co s co sff ffvvvθθ=−2, 2, 1, 1,sin sinffffvvθθ=1,0 1, 1,2,1, 1,coscotansinfffffvvvθθθ−=1, 1, 1,0cosffvvθ=21,0 1, 1,-12,1,0 1, 1,coscotancos
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