θ θ . Since this is a vector expressions we really have two equations that hold independently along each coordinate direction:

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1 Impulse, momentum (cont.) Top view A kg ball is thrown at a frictionless wall, at a 30 o x angle with incident speed of 6 m/s, as shown. After a 4.0 ms time of contact with the wall, θ θ 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: px = Fxav t & py = Fyav t y

2 Top view v yf ŷ v xi ˆx θ v yi θ v xf Along x: = = m m s s = o o px m(vxf v xi) m(6 cos30 6 cos30 ) 0 So no force acts along x (wall is frictionless).

3 Along y: py = m(vyf v yi ) m = s m s o o m 6 sin 30 6 sin 30 kg m py = 1.8 = Fyav t s Then solving for, F yav F yav = impulse on ball p kg m 1.8 s s y = = = t ŷ ˆx θ 450 N Top view v xi v yi θ v xf Average force on ball v yf The average force on the wall will be equal but opposite in direction to this so, = 450 N Fyav Average force on wall

A floating cooler had drifted to halfway between them, and they start a playful tug of war to get to

4 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

5 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 m 1 and m 2, the center of mass is a position along the line connecting the objects from which the distance to each object d 1 and d 2 is mass weighted such that: x cm m 1 m 2 m 1 d 1 = m 2 d 2 d 1 d 2

6 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). x com d m 1 d 2 1 m 2 m 1 d 1 = m 2 d 2 If we had m 1 = m 2 = m then the system would balance at d 1 = d 2 = d and the x com would lie midway between the objects. x com m d d m

7 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 x 1 and x 2. y x com m 1 m 2 x d 1 d 2 x 1 x 2

8 y x com m 1 m 2 x d 1 d 2 x 1 x 2 In this coordinate system, xcom = x1 + d1 d1 = xcom x1 x2 = xcom + d2 d2 = x2 xcom For the com, m 1 d 1 = m 2 d 2 substituting, rearranging, m 1(xcom x 1) = m 2(x2 x com ) mx 1 com mx 1 1 = mx 2 2 mx 2 com mx 1 com + mx 2 com = mx mx 2 2

9 (m1+ m 2)xcom = m1x1+ m2x2 Hence with m 1 at x 1 and m 2 at x 2 the com is located at, x com = mx m + mx + m If there were a third mass m 3 at x 3 we would find, x com = mx + mx + mx m + m + m And if there were n masses we would have, x n 1 = mx where, com i i M i = 1 n M= m i= 1 i

10 This only considers the masses distributed along the x axis, more generally, if the masses are distributed in space so that, mass m i is at position Then, r = xxˆ + yyˆ + zzˆ i i i i r = x xˆ + y yˆ + z zˆ com com com com (position of the com) where, x n 1 = mx com i i M i = 1 n 1, y = my, com i i M i = 1 z n 1 = mz com i i M i = 1 Which can also be written n 1 rcom = mr Vector eqn., gives the above 3 eqns., i i M i = 1 one along each coordinate axis.

11 Example: m 1 = 1 kg y m 2 = 2 kg r ˆ ˆ 2 = 2x + 1y r ˆ ˆ 1 = 2x + 0y x (meters) Then: 3 = i = + + = i= 1 M m 1kg 2 kg 1.5 kg 4.5 kg m 3 = 1.5 kg r ˆ ˆ 3 = 3x 2y n 1 1 rcom = m ˆ ir i = [{1kg( 2m) + 2kg(2m) + 1.5kg(3m)}x M i= kg + {1kg(0m) + 2kg(1m) + 1.5kg( 2m)}y] ˆ 1 r ˆ ˆ ˆ ˆ com = [{6.5kg m}x + { 1kg m}y] = (1.44m)x (0.22m)y 4.5 kg

12 m 1 = 1 kg r ˆ ˆ 1 = 2x + 0y y m 2 = 2 kg r ˆ ˆ 2 = 2x + 1y x com r ˆ ˆ com = (1.44m)x (0.22m)y m 3 = 1.5 kg r ˆ ˆ 3 = 3x 2y 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).

13 Newton s 2 nd law for a system of particles For a single object we have, F = ma net Consider the expression for the center of mass, r com n 1 = mr i i M i = 1 Rearranging and expanding, this can be written, Mr = m r + m r + m r m r com n n

14 Mr = m r + m r + m r m r com n n Allowing the motion to progress for a time t lets us write that, r r r r r M = m + m + m m t t t t t com n n 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 = m v + m v + m v m v com n n

15 Mv = m v + m v + m v m v com n n Using this trick again we allow the motion to progress for a time t letting us write that, vcom v1 v2 v3 vn M = m1 + m2 + m mn 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 = m a + m a + m a m a com n n

16 Ma = m a + m a + m a m a com n n The l.h.s. is the total mass of all the particles 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 = F + F + F F com n n M= m i= 1 i 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.

17 Now comes a key point which makes a critical distinction between these two types of forces: internal vs. external. Ma = F + F + F F com n All the internal forces can be written as pairwise forces, F n,m, which represents the that force particle n experiences due to particle m. But by Newton s 3 rd law, for every such force there must be an equal and opposite force F = F that particle m experiences due to n. m,n n,m This means that in our sum on the right all the internal forces will cancel each other out leaving only the external forces. Ma = F + F + F F com n

18 Hence Newton s 2 nd law for the system of particles is, Where, F net = Ma com F net = the sum of the external forces only. n M = the total mass of the system (i.e. M= mi ) a com= the acceleration of the center of mass of the system i= 1 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. F net

19 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, F = Ma = 0 net Which tells us that the center of mass does not move. com The only forces acting are therefore internal forces and we can write, Fxnet = Maxcom = 0= FxMaria + FxArnold F xmaria = F xarnold mmariaaxmaria = marnoldaxarnold

20 mmariaaxmaria = marnoldaxarnold But m Arnold = 2m Maria so, a a xmaria xmaria m = m 2m = m Arnold Maria maria Maria a a xarnold xarnold a = 2a So, xmaria xarnold Maria s acceleration (magnitude) is twice Arnold s.

21 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). Earth com It is this center of mass that revolves about the Sun in a slightly elliptical orbit. Sun Moon

22 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.

23 Example To measure the muzzle velocity of a very high velocity cartridge of mass m b, it is fired into a much more massive block (mass, m B ) 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 v f. What was the bullets muzzle velocity? v ib =? at rest, v ib = 0 v f bullet, m b Block, m B (m b +m B ) bullet & block Use conservation of momentum: Pf = Pi pfb + pfb = pib + pib

24 pfb + pfb = pib + pib 0 mv + mv = mv + mv b fb B fb b ib B ib & vfb = vfb = vf so, mv b f + mv B f = mv b ib (mb + m B)vf = mbvib v ib (mb + m B) = m b v f v ib =? at rest, v ib = 0 v f bullet, m b Block, m B (m b +m B ) bullet & block

25 Then if the bullet had mass m b =10 g and the block had mass m B =1 kg and their final speed was measured as 9 m/s, (m + m ) (0.01 kg kg) m b B vib = v f = (9.00 ) mb 0.01 kg s m vib = 101(9.00 ) s m vib = 909 s v ib =? at rest, v ib = 0 v f bullet, m b Block, m B (m b +m B ) bullet & block

26 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 v T v W m T + m C m T initial m C v C x

27 For the x direction: P xf = P xi 0 (m + m )v = m v + m v T C WX T TX C CX v P WX f = P i mt = v (m + m ) T (vector eqn) C TX v T m T initial m C y v C final v W m T + m C x For the y direction: P yf = P yi 0 (m + m )v = m v + m v T C Wy T Ty C Cy v Wy = mc (m + m ) T C v Cy

28 Then, m m v = v xˆ + v yˆ T C W Tx Cy mt + mc mt + mc m m v ˆ ˆ W = 3.57 x y s s Which has angle (w.r.t. x axis), v T initial m T m C y v C final v W m T + m C x 2.86 θ= Tan = And magnitude, 1 o v W m θ 2.86 s m 3.57 s W ( ) ( ) 2 2 v = 3.57 m / s m / s = 4.57 m / s

29 No external forces acted in the collision in the plane of the motion. This means that, Fnet = Macom = 0 vcom y a final com = = 0 t v Which implies that W v T m T + m C vcom = 0 m x T v v = 0 f com v i com f com = v i com initial 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. m C v C

Quiz Samples for Chapter 9 Center of Mass and Linear Momentum

Quiz Samples for Chapter 9 Center of Mass and Linear Momentum Name: Department: Student ID #: Notice +2 ( 1) points per correct (incorrect) answer No penalty for an unanswered question Fill the blank ( ) with ( ) if the statement is correct (incorrect) : corrections