Physics 2210 Fall smartphysics Exam 3 Review smartphysics units /04/2015

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1 Physics 22 Fall 25 smartphysics Exam 3 Review smartphysics units -3 /4/25

2 Review Problem The figure shown extends from x = to x = and is bounded on the left by the y-axis, on the bottom by the x-axis, and on top by the curve y = ae x b y = ae x b. It has h uniform density ρ and thickness t (in the z direction). Find the x- and y-coordinates of its center of mass. dd We first have to find the mass of this figure. It is finite even though its width is infinite. We divide the shape into vertical strips. The mass element here is then dd = ρ dd = ρρ dd = ρρρ dd x b Where dd is the width of the strip and h = ae M = dd = ρρρ dd = ρρρ e is the height of each strip. = ρρρ b e u du x b dd Where we have made the substitution u = x b, and du = b dd M = ρρρρ e u y a = ρρρρ = ρρρρ x The x-and y-coordinates of the center-of-mass of the figure, X CC, Y CC are given by: X CC = M x CCdd, Y CC = M y CCdd Where (x CC, y CC ) = x, h 2 is the center-of mass of each vertical strip.

3 Problem (continued) x b y = ae : x = to x = ; uniform density ρ and thickness t. Find the x- and y-coordinates of its center of mass. We first solve for X CC X CC = M x CCdd = ρρρρ x ρρρ dd = ρρρρ b ρρρ x e x dd Again let u = x x b b, but also v = e = e u, and we integrate by parts: X CC = u= ρρρρ ρρρ b b u eu dd Note that dv = d e u = e u dd, and so we have X CC = u= ρρρρ ρρρb2 u dv = b uu u= u= u= u = u = b ue u = eu dd u= = ρb X CC = b u = u = u= v du u= = b e u

4 Problem (continued) x b y = ae : x = to x = ; uniform density ρ and thickness t. Find the x- and y-coordinates of its center of mass. Next for Y CC Y CC = M y CCdd = ρρρρ h 2 ρρρ dd = a e 2x b dd 2b Again let w = 2x b dw = 2 b dd Y CC = a 2b b 2 ew dd = ρρρρ ρρ 2 a2 x b e x b e dd = a 4 ew = a 4 Y CC = a 4

5 Problem (continued) Alternate solution for Y CC : Divide figure into horizontal strips of height dd and width w = x(y). Invert relationship between x and y of curved border: x b y = ae x b e = y a x b = ln y a x = b ln y a And so the mass element is given by dd = ρρρ dd = ρρρ ln y dd a Y CC = M y CCdd = a ρρρρ ρρρ y ln y a dd = a a u ln u du = a a 4 dy y a w x = b ln y a x Y CC = a 4 Where we have made the substitution u = y a dd = aaa, We evaluate u ln u du by making the substitution v = u ln u u u ln u = v + u and dd = ln u dd, u= u ln u du = udd = uu u = u= u = vdu = u 2 ln u u 2 u= u ln u u dd u= u= u= u= = u ln u du + uuu = u ln u du + 2 u2 = u ln u du + 2 = u ln u du 2 2 u ln u du = 2 u= u ln u du = 4

6 Review Problem 2 (/4) A blue car of mass m =7kg was initially traveling east at v i =27.m/s. A red car of mass m 2 = kg was traveling north at v 2i =43.m/s. They collide at an intersection, and (kinetic) energy was lost. After the collision, the blue car is now traveling at φ = 5 north of east, and the red car at φ 2 = 2 north of east. (a) Find speeds v and v 2f after the collision. (b) How much energy was lost? y (north) Solution: (a) No external forces act horizontally on the system of two cars, so the horizontal components of their total momentum is conserved in the collision, even if energy is not. Before collision: P ii = m v i, P iy = m 2 v 2i After the collision we then have (using conservation of momentum): P fx = m v f cos φ + m 2 v 2f cos φ 2 P fy = m v f sin φ + m 2 v 2f sin φ 2 And so by conservation of momentum we have m v f cos φ + m 2 v 2f cos φ 2 = m v i () m v f sin φ + m 2 v 2f sin φ 2 = m 2 v 2i (2) m v i v 2i m 2 v f v 2f φ = 5 φ 2 = 2 x (east)

7 Review Problem 2 (2/4) blue car m =7kg east at v i =27.m/s. red car of mass m 2 = kg north at v 2i =43.m/s. Inelastic collision: after: blue car at φ = 5 north of east, and the red car at φ 2 = 2 north of east. (a) Find speeds v and v 2f after the collision. (b) How much energy was lost? (a) continued: m v f cos φ + m 2 v 2f cos φ 2 = m v i () m v f sin φ + m 2 v 2f sin φ 2 = m 2 v 2i (2) Taking sin φ 2 cos φ 2 2 : m v f cos φ sin φ 2 m v f sin φ cos φ 2 = m v i sin φ 2 m 2 v 2i cos φ 2 v f = m v i sin φ 2 m 2 v 2i cos φ 2 m (cos φ sin φ 2 sin φ cos φ 2 ) = m v i sin φ 2 m 2 v 2i cos φ 2 m sin( φ 2 φ ) = 7kg 27.m/s sin 2 kg 43.m/s cos 2 7kg sin( 3 ) v f = 33.8m/s

8 Review Problem 2 (3/4) blue car m =7kg east at v i =27.m/s. red car of mass m 2 = kg north at v 2i =43.m/s. Inelastic collision: after: blue car at φ = 5 north of east, and the red car at φ 2 = 2 north of east. (a) Find speeds v and v 2f after the collision. (b) How much energy was lost? (a) continued: m v f cos φ + m 2 v 2f cos φ 2 = m v i () m v f sin φ + m 2 v 2f sin φ 2 = m 2 v 2i (2) Taking sin φ cos φ 2 : m 2 v 2f cos φ 2 sin φ m 2 v 2f sin φ 2 cos φ = m v i sin φ m 2 v 2i cos φ v 2f = m v i sin φ m 2 v 2i cos φ m 2 (cos φ 2 sin φ sin φ 2 cos φ ) = m v i sin φ m 2 v 2i cos φ m 2 sin( φ φ 2 ) = 7kg 27.m/s sin 5 kg 43.m/s cos 5 kg sin 3 v 2f = 8.65m/s

9 Review Problem 2 (4/4) blue car m =7kg east at v i =27.m/s. red car of mass m 2 = kg north at v 2i =43.m/s. Inelastic collision: after: blue car at φ = 5 north of east, and the red car at φ 2 = 2 north of east. (a) Find speeds v f and v 2f after the collision. (b) How much energy was lost? (b) E LLLL = K = K i K f K i = 2 m v 2 i + 2 m 2v 2 2i = 7kg 27.m/s 2 + kg 43.m/s 2 =.64 6 J 2 K f = 2 m v 2 f + 2 m 2v 2 2i = 7kg 33.8m/s 2 + kg 8.65m/s 2 =. 6 J 2 E LLLL = K = K i K f =.64 6 J. 6 J E LLLL = J

10 Review Example 3 (/3) (hr9-65) A body of mass 2. kg makes a HEAD-ON elastic collision with another body at rest and continues to move in the original direction but with one-fourth of its original speed. (a) What is the mass of the other body? (b) What is the speed of the two-body center of mass if the initial speed of the 2. kg body was 4. m/s? Solution: We denote the 2kg mass as body and the other as body 2 m = 2.kg, v f = 4 v i part a m 2 =?, v 2i = (a) Elastic Collision: So we first find V CC (in this case without assuming values) V CC = m v i + m 2 v 2i = m v i m + m 2 m + m 2 Transform velocity of body into the CM frame v i = v i V CC = v i m v i = m v i + m 2 v i m v i = m 2v i m + m 2 m + m 2 m + m 2 Elastic collision in D: v f = v i= m 2v i m + m 2

11 Review Example 3 (2/3) (hr9-65) A body of mass 2. kg makes a HEAD-ON elastic collision with another body at rest and continues to move in the original direction but with one-fourth of its original speed. (a) What is the mass of the other body? (b) What is the speed of the two-body center of mass if the initial speed of the 2. kg body was 4. m/s? (a) continued Transform back into Lab frame v f = m 2v i m + m 2 v f = v f + V CC = m 2v i + m v i = m m 2 v m + m 2 m + m 2 m + m i 2 But we were given v f = 4 v i, m m 2 m + m 2 = 4 4m 4m 2 = m + m 2 3m = 5m 2 m 2 = 3 5 m = kg m 2 =.2kg

12 Review Example 3 (3/3) (hr9-65) A body of mass 2. kg makes a HEAD-ON elastic collision with another body at rest and continues to move in the original direction but with one-fourth of its original speed. (a) What is the mass of the other body? (b) What is the speed of the two-body center of mass if the initial speed of the 2. kg body was 4. m/s? (b) Now given: v i =4. m/s We had from part (a) m 2 =.2 kg Again: the velocity of the center of mass is V CC = m v i m + m 2 2.kg 4. m s = 2.kg +.2kg V CC = 2.5m/s

13 Extra Review Example 4 (/3) Electrical interaction between two protons can be expressed as a potential energy: U = +k e 2 r, where e =.6 9 coulomb (C) is the charge of each proton (a.k.a. elementary charge ), k = N m 2 /C 2, and r is the distance between the two protons. A proton moving at v = m/s collides elastically head-on with a second proton moving at v 3 =.5 7 m/s. Protons have mass m = kg. The full process occurs strictly along the x-axis. Find (a) The final velocity of the second proton, and (b) The closest distance of approach of between the two protons. Solution: Since the interaction can be written as a potential energy it is conservative and so the collision is elastic. First we find the center of mass velocity: V CC = m v i + m 2 v 2i = mv + m( v 3) = + v m + m 2 2m 3 Transforming into the center-of-mass frame: we are concerned only with the second proton vv 2i = v 2i V CC = v 3 + v 3 = 2v 3 The collision is elastic, so the second proton reverses direction: vv 2f = v 2i = + 2v 3 Transforming back into the Lab Frame: vv 2f = vv 2f + V CC = + 2v v 3 = v = m/s

14 Extra Example 4 continued two protons U = +k e 2 r, e =.6 9 C, k = N m 2 /C 2, and r = distance between the two protons. Proton at v = m/s collides elastically head-on with proton 2 at v 3 =.5 7 m/s. Proton mass m = kg. Find (a) The final velocity of the second proton, and (b) The closest distance of approach of between the two protons. At closest approach in a D collision: x 2 x is a minimum, which means d dd x 2 x v 2 v = v 2 = v No net external force acts on the system, so V CC must be constant, so at closest approach we must have v 2 = v = V CC. And so at this moment then v 2 = v = + v 3. Electrical interaction is conservative (because it can be written as a potential energy): Total energy is conserved. The total energy is given by E = 2 m v m 2v kk2 r Before collision (assume the separation is infinite, so k e 2 r = ) At minimum separation: r = r mmm E = E = 2 m v m v 3 E = E 2 = 2 m + v m + v 3 2 = 5 9 mv kk2 r mmm = 9 mv 2

15 Extra Example 4 continued two protons U = +k e 2 r, e =.6 9 C, k = N m 2 /C 2, and r = distance between the two protons. Proton at v = m/s collides elastically head-on with proton 2 at v 3 =.5 7 m/s. Proton mass m = kg. Find (a) The final velocity of the second proton, and (b) The closest distance of approach of between the two protons. Setting E = E 2 by conservation of total energy: 9 mv 2 + kk2 = 5 r mmm 9 mv 2 kk 2 = 4 r mmm 9 mv 2 r mmm = 9kk2 4mv 2 r mmm = N m 2 /C C kg.5 7 m/s 2 r mmm =.38 5 m