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Transcript
Impulse, momentum (cont.) y Top view A 0.300 kg ball is thrown x at a frictionless wall, at a 30o θ θ angle with incident speed of 6 m/s, as shown. After a 4.0 ms time of contact with the wall, the ball comes off with a numerically equal angle, at the same speed. What is the impulse on the ball and average force experienced by the wall during the collision? For the ball ∆p = Fav ∆t . Since this is a vector expressions we really have two equations that hold independently along each coordinate direction: ∆p x = Fxav ∆t ∆p y = Fyav ∆t & Top view vyf vxi vyi ŷ x̂ θ vxf θ Along x: m(v xf − v xi= ) m(6 ∆p= x m m o cos30o − 6 cos30= ) 0 s s So no force acts along x (wall is frictionless). Top view Along y: ∆p= m(v yf − v yi ) y vyf vxi vyi ŷ x̂ m m o o m 6 sin 30 − −6 sin 30 s s θ vxf θ kg m ∆p y = 1.8 = Fyav ∆t = impulse on ball s Then solving for Fyav , ∆p y F = = yav ∆t kg m s= 450 N 0.004 s 1.8 Average force on ball The average force on the wall will be equal but opposite in direction to this so, Average force Fyav = −450 N on wall HITT In better days Maria & Arnold were on their honeymoon afloat on separate inner tubes, tied together by a 10 m rope. Arnold was 2 times heavier and about 4 times stronger. A floating cooler had drifted to halfway between them, and they start a playful tug of war to get to the cooler. Who gets to the cooler first? Philanthropist & ex-wife of Ex-governor of CA Ex-governor of CA A) Arnold B) Maria C) Its a tie Maria wins, even if she merely holds on with one hand and Arnold does all the pulling. No matter how hard one or the other (or both) pull, they will meet at their center of mass. We’ll soon see why this happens. For now we just want to define the center of mass and learn how to calculated it in various circumstances. For two objects of mass m1 and m2, the center of mass is a position along the line connecting the objects from which the distance to each object d1 and d2 is mass weighted such that: xcm m1 d1 d2 m2 m1 d1 = m2 d2 If the masses were connected by a rigid rod and the system suspended by a knife edge support, the com is the point along the rod at which the system will balance (also called the center of gravity but only the same point in a uniform gravitational field). xcom m1 d1 d2 m2 m1 d1 = m2 d2 If we had m1= m2= m then the system would balance at d1= d2 = d and the xcom would lie midway between the objects. xcom m d d m Usually we need to locate the objects and the com within a coordinate system, hence suppose the two masses lie along the x axis at positions x1 and x2. y xcom m1 x1 d1 d2 m2 x2 x y xcom m1 d1 x1 In this coordinate system, x com= x1 + d1 = x 2 x com + d 2 d2 m2 x2 = d1 x com − x1 d= x 2 − x com 2 For the com, m1 d1 = m2 d2 substituting, m1 (x com − x1= ) m 2 (x 2 − x com ) rearranging, m1x com − m1x1 = m 2 x 2 − m 2 x com m1x com + m 2 x com = m1x1 + m 2 x 2 x (m1 + m 2 )x com = m1x1 + m 2 x 2 Hence with m1 at x1 and m2 at x2 the com is located at, x com m1x1 + m 2 x 2 = m1 + m 2 If there were a third mass m3 at x3 we would find, x com m1x1 + m 2 x 2 + m 3x 3 = m1 + m 2 + m 3 And if there were n masses we would have, x com 1 n = ∑ mixi M i =1 n where, M = ∑ mi i =1 This only considers the masses distributed along the x axis, more generally, if the masses are distributed in space so that, mass mi is at position ri = x i xˆ + y i yˆ + z i zˆ Then, rcom = x com xˆ + y com yˆ + z com zˆ where, 1 n x com = ∑ m i x i , M i =1 y com (position of the com) 1 n 1 n = ∑ m i y i , z com = ∑ m i z i M i =1 M i =1 Which can also be written 1 n rcom = ∑ m i ri M i =1 Vector eqn., gives the above 3 eqns., one along each coordinate axis. y Example: m1= 1 kg r1 = −2xˆ + 0yˆ Then: 3 M =∑ m i =1kg + 2 kg + 1.5 kg =4.5 kg m2= 2 kg = r2 2xˆ + 1yˆ x (meters) m3= 1.5 kg = r3 3xˆ − 2yˆ i =1 rcom = = rcom 1 n 1 m i ri [{1kg( −2m) + 2kg(2m) + 1.5kg(3m)}xˆ = ∑ M i =1 4.5 kg ˆ +{1kg(0m) + 2kg(1m) + 1.5kg( −2m)}y] 1 ˆ (1.44m)xˆ − (0.22m)yˆ [{6.5kg m}xˆ + {−1kg = m}y] 4.5 kg y m1= 1 kg m2= 2 kg = r2 2xˆ + 1yˆ x com r1 = −2xˆ + 0yˆ = rcom (1.44m)xˆ − (0.22m)yˆ m3= 1.5 kg = r3 3xˆ − 2yˆ Notice that the objects here have different shapes. This doesn’t matter as long as their positions were given in terms of the positions of their individual coms (which for symmetric objects, of uniform density, are at their centers). Newton’s 2nd law for a system of particles For a single object we have, Fnet = ma Consider the expression for the center of mass, 1 n rcom = ∑ m i ri M i =1 Rearranging and expanding, this can be written, Mrcom= m1 r1 + m 2 r2 + m 3 r3 + ... + m n rn Mrcom= m1 r1 + m 2 r2 + m 3 r3 + ... + m n rn Allowing the motion to progress for a time ∆t lets us write that, ∆rcom ∆r3 ∆r1 ∆r2 ∆rn M = m1 + m2 + m3 + ... + m n ∆t ∆t ∆t ∆t ∆t Recall that in the limit of the time interval going to zero ∆r lim = v (instantaneous velocity) ∆t →0 ∆t In that limit we get that, Mv com= m1v1 + m 2 v 2 + m 3 v 3 + ... + m n v n Mv com= m1v1 + m 2 v 2 + m 3 v 3 + ... + m n v n Using this trick again we allow the motion to progress for a time ∆t letting us write that, ∆v com ∆v 3 ∆v1 ∆v 2 ∆v n M = m1 + m2 + m3 + ... + m n ∆t ∆t ∆t ∆t ∆t Now in the limit of the time interval going to zero ∆v lim = a (instantaneous acceleration) ∆t →0 ∆t Taking that limit again we now get that, Ma com= m1a1 + m 2a 2 + m 3a 3 + ... + m n a n Ma com= m1a1 + m 2a 2 + m 3a 3 + ... + m n a n n The l.h.s. is the total mass of all the particles M = ∑ m i i =1 times the acceleration of the com of the system of particles. While the r.h.s. we can recognize as the sum of the individual forces acting on the individual particles, Ma com = F1 + F2 + F3 + ... + Fn Included in these forces on the right are all the external forces acting on the system of particles as well as the internal forces that the particles exert on one another. Now comes a key point which makes a critical distinction between these two types of forces: internal vs. external. Ma com = F1 + F2 + F3 + ... + Fn All the internal forces can be written as pairwise forces, Fn,m , which represents the that force particle n experiences due to particle m. But by Newton’s 3rd law, for every such force there must be an equal and opposite force Fm,n = − Fn,m that particle m experiences due to n. This means that in our sum on the right all the internal forces will cancel each other out leaving only the external forces. Ma com = F1 + F2 + F3 + ... + Fn Hence Newton’s 2nd law for the system of particles is, Fnet = M a com Where, Fnet = the sum of the external forces only. n M = the total mass of the system (i.e. M = ∑ m i ) i =1 a com = the acceleration of the center of mass of the system Note that the com behaves like it is a single point object of mass M (total mass of all the objects) acted upon by the external force Fnet . Returning to Maria and Arnold, and their tug of war, we can now see why Maria will always gets to the cooler first. Since no external forces act, = Fnet M = a com 0 Which tells us that the center of mass does not move. The only forces acting are therefore internal forces and we can write, Fxnet= M a xcom= 0= FxMaria + FxArnold FxMaria = − FxArnold m Maria a xMaria = − m Arnold a xArnold m Maria a xMaria = − m Arnold a xArnold a xMaria m Arnold = − a xArnold m Maria But mArnold = 2mMaria so, a xMaria So, 2m maria = − a xArnold m Maria a xMaria = −2a xArnold Maria’s acceleration (magnitude) is twice Arnold’s. We speak of the Moon revolving about the Earth, but this is slightly inaccurate. Both the Earth and the Moon revolve around their common com. Because the Earth is so much more massive than the Moon their center of mass actually lies within the volume of the Earth (~2/3 of the way from the center of the Earth). It is this center of mass that revolves about the Sun in a slightly elliptical orbit. Earth com Moon Sun Collisions Types: Completely elastic: kinetic energy is conserved, at least during the time of the collision (idealization). Completely inelastic: energy dissipated in the collision and bodies stick together after the collision. In line (direct hit): all motion (before and after along same line) General: use symmetry to advantage. Example To measure the muzzle velocity of a very high velocity cartridge of mass mb, it is fired into a much more massive block (mass, mB) that is initially stationary on an air track. The bullet embeds itself in the block and the more manageable speed of the two moving off after the collision is measured to be vf. What was the bullets muzzle velocity? vib = ? bullet, mb at rest, viB = 0 Block, mB Use conservation of momentum: vf (mb +mB) bullet & block Pf = Pi pfb + pfB = pib + piB pfb + pfB = pib + piB 0 m b v fb + m B v fB = m b v ib + m B v iB v= v f so, & v= fb fB m b vf + m B vf = m b v ib (m b + m B )v f = m b v ib (m b + m B ) v ib = vf mb vib = ? bullet, mb at rest, viB = 0 Block, mB vf (mb +mB) bullet & block Then if the bullet had mass mb =10 g and the block had mass mB =1 kg and their final speed was measured as 9 m/s, v ib (m b + m B ) (0.01 kg + 1.00 kg) m vf (9.00 ) = mb 0.01 kg s v ib = 101(9.00 v ib = 909 vib = ? bullet, mb m ) s m s at rest, viB = 0 Block, mB vf (mb +mB) bullet & block Example On a completely iced road a car of mass 1000 kg, traveling north through an intersection at 36 km/hr (10 m/s), collides with a pick-up truck of mass 2500 kg, traveling east at half that speed. The two vehicles lock together after the collision. What is the direction of the wreck and its velocity immediately after the collision? y final vW mT+ mC vT mT initial m C x vC Pf = Pi (vector eqn) y final For the x direction: Pxf = Pxi (m T + m C )v WX v WX vW mT+ mC vT 0 =m T v TX + m C v CX mT initial m mT v TX = (m T + m C ) C For the y direction: Pyf = Pyi (m T + m C )v Wy v Wy 0 =m T v Ty + m C v Cy mC = v Cy (m T + m C ) x vC Then, vW vW y mC mT ˆ v Tx x + v Cy yˆ mT + mC mT + mC m m ˆ 3.57 x + 2.86 yˆ s s Which has angle (w.r.t. x axis), final vW mT+ mC vT x mT initial 2.86 o θ Tan = 38.7 3.57 vC mC −1 vW θ m 3.57 s And magnitude, vW = ( 3.57 m / s ) + ( 2.86 m / s ) 2 2 m 2.86 s = 4.57 m / s No external forces acted in the collision in the plane of the motion. This means that, = Fnet M = a com 0 ∆v com y final = a com = 0 ∆t vW Which implies that mT+ mC vT ∆v com = 0 x m T v f com − v i com = 0 initial v f com = v i com mC vC Which means that whatever the vector velocity of the com before the collision, the com of the system maintains that same vector velocity after the collision. This would be true even if the vehicles separated after the collision.