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7.1 the Impulse-Momentum Theorem

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7.1 the Impulse-Momentum Theorem 7.1 The Impulse-Momentum Theorem TRSP Fig. 7-1b: Force on a baseball. Definition of Impulse: the impulse of a force is the product of the average force F and the time interval ∆t during which the force acts: Impulse = F ∆t (7.1) Impulse is a vector quantity and has the same direction as the average force. SI Unit of Impulse: newton second ( N s) · · Definition of Linear Momentum: the linear momen- tum −→p of an object is the product of the object’s mass m and velocity −→v : −→p = m−→v (7.2) TRSP Fig. 7.4: average force and velocity change v v0 −→f −−→ a = ∆t v v0 mv mv0 −→f −−→ −→f −−−→ From N2: F = m ∆t = ∆t (7.3) Ã ! Impulse-Momentum Theorem: When a net force acts on an object, the impulse of the net force is equal to the change in momentum of the object: F ∆t = mv mv (7.4) −→f − −→0 Impulse = Change in momentum Ex.1:Hittingabaseball(mass,m =0.14 kg), initially −v→0 = 38 m/ s, −v→f =58m/ s, ∆t =1.6 3 − × 10− s (a) impulse = mv mv −→f − −→0 m =0.14 kg [58 m/ s ( 38 m/ s)] = . 14 ( kg) 96.0 s = m − − 13. 44 ( kg) s m impulse 13. 44( kg) s m (b) F = ∆t = 3 = 8400.0(kg) 2 or 1.6 10− s s N × Ex.2: Rain Storm: v = 15 m/ s, rate of rain is 0.060 kg/ s. Find aver- −→0 − ageforceonthecarifraindropscometorest. average force on the rain m−v→f −m−v→0 m F = − = v0 ∆t − ∆t −→ ³ ´ =0.060 kg/ s 15 m/ s= . 9(kg) mor N × − − s m Force on roof = +. 9(kg) s or N CEx. 3: Hailstones versus Raindrops: Hailstones re- bound giving a greater force. Problems 8, 10. 7.2 The Principle of Conservation of Linear Momentum TRSP Fig. 7.7 Concepts at a Glance On page 195/6 Cutnell and Johnson prove that for an isolated system (no external force, e.g. gravity): −P→f = −P→i (7.7) where −P→f = m1−v→f1 + m2−v→f2, and −P→i = m1−v→i1 + m2−v→i2 Principle of Conservation of Linear Momentum: The total linear momentum of an isolated system remains constant (is conserved). An isolated system is one for which the vector sum of the external forces acting on the system is zero. CEx. 4: READ! Ex. 5: Assembling a Freight Train: 3 m1 =65 10 kg, v01 =+0.80 m/ sisovertakenby × 3 m2 =92 10 kg, v01 =+1.20 m/ s × Find the common velocity when they are coupled: (m1 + m2) vf = m1v01 + m2v02 m1v01+m2v02 which solves for vf = 1.0m/ s m1+m2 → Ex.6: Ice Skaters: TRSP Fig. 7.11: simply a case of total momentum being zero - READ [CJ4: CEx. 5. Firing Blanks] Problems 20, 46 REASONING STRATEGY: page 200 - READ 7.3 Collisions in One Dimension Elastic collisions Inelastic collisions TRSP Fig. 7.13 Ex. 7: 1D collision: m1 =0.250 kg, v01 =5.00 m/ s collides head-on with m2 =0.800 kg, v02 =0(at rest): Fig. 7.12 (DRAW) m1vf1 + m2vf2 = m1v01 +0 (A) 1 2 1 2 1 2 2m1vf1 + 2m2vf2 = 2m1v01 +0 (B) Generalization of 1D to include an initially moving m2: m1vf1 + m2vf2 = m1v01 + m2v02 (A’) 1 2 1 2 1 2 1 2 2m1vf1 + 2m2vf2 = 2m1v01 + 2m2v02 (B’) SIMPLIFY THE VARIABLES: m1 = a, m2 = b, vf1 = x, vf2 = y, v01 = c, v02 = d: Solve A’: ax + by = ac + bd, Solution is : y = xa ac bd , − − b − n o eliminate y in B’ after dropping the ”halves”: ax2 + by2 = ac2 + bd2 gives a quadratic in x: 2 ax2 + b xa ac bd = ac2 + bd2, − − b − ³ ´ Solutions are : x = c , { } 2bd+ac cb and x = b+a− n o The first solution is incompatible with a moving m2 after the collision, hence we choose the second solu- tion to substitute into the momentum equation, A’: ax + by = ac a 2bd+ac cb + by = ac, → b+a− ³ ´ d+c Solution is : y =2a−b+a , n o NOW set d = v02 =0: m1 m2 Thus v = − v (7.8a) f1 m1+m2 01 ³ ´ and v = 2m1 v (7.8b) f2 m1+m2 01 ³ ´ Using the numerical values: v = 2.62 m/ s(soball f1 − 1 rebounds left) and vf2 =+2.38 m/ s N.B. if m1 <m2 then vf1 will be negative. ALSO if m1 = m2 then vf1 will be zero (stationary). Ex. 8: Ballistic Pendulum: Fig. 7.15 (DRAW) bullet: m1 =0.01 kg , speed v01;blockofwood: m2 =2.5kgrisesto h =0.65 m: Find v01. conserve momentum: (m1 + m2) vf = m1v01(a), v Solution is : v =(m + m ) f 01 1 2 m1 n o 1 2 conserve energy: (m1 + m2) ghf = 2 (m1 + m2) vf (b), CANCEL common factor, so that: 2 vf =2ghf Numerically: vf =3.57 m/ sandv01 = 896 m/ s. 7.4 Collisions in Two Dimensions TRSP Fig. 7.16 Ex. 9: 2D collision: DATA IN FIG. 7.15: x component: Pfx = P0x (7.9a) (15 kg) vf1x +(26kg)(0.7m/ s) (cos 35 ◦)= ³ ´ (15 kg) (0.9m/ s) (sin 50 ◦)+(26kg)(0.54 m/ s) v =( 14. 909 + 24. 382) /15.0=. 631 53 m/ s f1x − y component: Pfy = P0x (7.9b) (15 kg) v +(26kg)( 0.7m/ s) (sin 35 )= f1y − ◦ ³ ´ (15 kg) ( 0.9m/ s) (cos 50 )+0 − ◦ v =( 10. 439 + 8. 677 6) /15.0=. 117 43 m/ s f1y − The final speed of m1 is: 1 2 2 2 vf1 = . 631 53 + . 117 43 = . 642 35 m/ s ³ ´ anditsdirectionhas: . 117 43 θ =arctan . 631 53 = . 183 85 rad or 10.5 ◦ ³ ´ 7.7 Centre of Mass (abbreviate: CM) x = m1x1+m2x2+... (7.10) CM m1+m2+... For 2 masses, if m1 = m2 = m , 1 then xCM = 2 (x1 + x2). Symmetric objects have the CM at their geometric centres. Velocity of the CM: ∆x = m1∆x1+m2∆x2 CM m1+m2 divide by ∆t: v = m1v1+m2v2 (7.11) CM m1+m2.
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