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Momentum and impulse • The linear momentum - p of an object of mass m moving with velocity v is the product of its mass and velocity: p = mv SI unit: kg m/s - Doubling the mass of the velocity of an object double its momentum - Doubling both quantities, quadruples its momentum Momentum is a vector quantity ( quiz 6.1/161) • Changing the momentum of an object require application of a force: Fnet = ma = m Δv/Δt = Δ(mv)/Δt The change in an object’s momentum Δp divided by the elapsed time Δt equals the constant net force Fnet acting on the object If a constant force acts on a object. The impulse I delivered to the object over a time interval Δt is given by: I = F Δt SI unit: kg m/s (ex 6.2/163) • Conservation of momentum • A2 /188 • When a collision occurs in an isolated system, the total momentum of the system doesn’t change with the passage of time • The total momentum is conserved! • If we consider an isolated system (no external forces) of two particles before and after they collide F21Δt = m1v1f-m1v1i F12 Δt = m2v2f-m2v2i F21= -F12 F21 Δt= -F12 Δt m1v1f-m1v1i = -(m2v2f-m2v2i) m1v1i +m2v2i= m1v1f+m2v2f When no net forces acts on the system, the total momentum remains ct. in time (conservation of momentum) • Collisions • For any type of collision, the momentum of the system just before the collision= to the momentum after the collision as long as the system is isolated • Elastic collision- both momentum and kinetic energy are conserved • Inelastic collision- momentum is conserved but kinetic energy is not • Perfect inelastic collision, momentum is conserved, KE is not, and the 2 objects stick together after the collision (quiz 6.3/168) • Perfectly inelastic collision: m1v1i +m2v2i =(m1+m2)vf vf=(m1v1i +m2v2i )/(m1+m2) (quiz 6.6/172) • Elastic collision: m1v1i +m2v2i= m1v1f+m2v2f m1(v1i-v1f) = m2(v2f-v2i) 1/2m1v1i2+1/2 m2v2i2=1/2 m1v1f2=1/2m2v2f2 m1(v1i2-v1f2)=m2(v2f2-v2i2) m1(v1i-v1f)(v1i+v1f)= m2(v2fv2i)(v2f+v2i) v1i+v1f =v2f+v2i v1i-v2i = -(v1f-v2f) • One dimensional collision; pb. Strategies - Choose coordinate axis (along direction of motion) - Diagram - Conservation of momentum: write total momentum before and after - Conservation of energy: write the expression (for elastic or perfect inelastic) - Solve (ex. Page 182-188) Glancing collisions • For a general collision in 3D, the conservation of momentum principle implies that total momentum of the system in each direction is conserved m1v1ix +m2v2x= m1v1fx+m2v2fx m1v1i y+m2v2iy= m1v1fy+m2v2fy Glancing collision: if an object with mass m1 collides with a mass m2 that is initial at rest, after collision, obj.1 moves at an angleθ with horizontal, and obj.2 moves at an angleΦ with horizontal x component: m1v1i + 0 = m1v1f cosθ+m2v2f cosΦ y component: 0 + 0 = m1v1f sinθ+m2v2f sinΦ Conservation of energy: 1/2m1v1i2+0 =1/2 m1v1f2+1/2m2v2f2 (if the collision is inelastic, the KE of the system is not conserved, the 3rd equation does not apply) • Two dimensional collision; pb. Strategies - Choose coordinate axis (use both x, and y- coordinates) - Diagram - Conservation of momentum: write total momentum before and after - Conservation of energy: write the expression (for elastic or perfect inelastic) - Solve Rocket Propulsion (page 177) • Equating the total initial momentum of the system with the total final momentum: (M + Δm)v =M(v+ Δv) + Δm(v-ve) ve –fuel ejected speed M Δv = - ve Δm but Δm =- ΔM M Δv = - ve ΔM (calculus:) vf-vi =veln(Mi/Mf) • The thrust of the rocket = the force everted on the rocket by the ejected exhaust gases instantaneous thrust= Ma = M Δv/ Δt = |ve ΔM/ Δt|