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Step 3: Motion in a Vertical Plane y = 1/2 gt 2

2. 1. Step 2: Elastic collision between the two masses Mass 1 collides with Mass 2 m 2 = 0.5m 1 m 1 v 1 + m 2 v 2 = m 1 v 1 1 + m 2 v 2 1 m 1 (2.42)+ 0= m 1 v 1 1 + 0.5m 1 v 2 1 2.42 = v 1 1 + 0.5v 2 1.

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Step 3: Motion in a Vertical Plane y = 1/2 gt 2

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  1. 2 1 Step 2: Elastic collision between the two masses Mass 1 collides with Mass 2 m2 = 0.5m1 m1v1 + m2v2 = m1v11 + m2v21 m1(2.42)+ 0= m1v11 + 0.5m1v21 2.42 = v11 + 0.5v21 v1 - v2 = v21 - v11 v21 - v11 2.42 - 0 = 2.42 = v21 - v11 Solve the two equations v11 = 0.81 m/s v21 = 3.23 m/s KE1 = 1/2 m1v12 = .5 m1v12 KE2 = 1/2(1.5 m1)(.67v1)2 = .34 m1v12 KE1 = .5 m1v12 34. m2 = 1.5 m1 m1v1 + m2v2 = 0 m1v1 + 1.5m1v2 = 0 v1 = - 1.5v2 v2 = - v1/1.5 v2 = - .67v1 KE2= .34 m1v12 m1v12 = 8928.57 KE1 + KE2 = 7500 .5 m1v12 + .34 m1v12 KE1 = .5 (8928.57) KE1 = 4464.29 J 4.46 kJ KE2= .34(8928.57) KE2 = 3035.71 J 3.04 kJ = 7500 m1v12 = 8928.57 1 page 189 1 29. Step 1: Mass 1 goes down ramp PE = KE mgh = 1/2 m v2 9.8(.3) = 1/2 v2 v = 2.42 m/s 1 Step 3: Motion in a Vertical Plane y = 1/2 gt2 0.9 = 1/2 (9.8) t2 t = 0.43 seconds 2 x = v1t x = 0.81(0.43) x = 0.35m 35cm for Mass 1 x = 3.23(0.43) x = 1.39m for Mass 2 # 32 on next page!!!!

  2. Step 2 L θ L - h x h v1 + m M M m + m M Conservation of Energy KE = PE 1/2 (m+M)(v1)2 = (m+M)gh (v1)2 = 2(m+M)gh (m+M) √2(9.8)h (v1)2 = 2gh v1 = √2gh page 189 32 Read example 7-10 for formulas these are used for a ballistic pendulum you must know how to do this! Step 1 L v Conservation of Momentum mv = (m+M)v1 For Ballistic Pendulum v = m+M √2gh ( ( v = m+M (v1) m 230 = .028+3.6 ( m ( .028 h = .16 m If you can't memorize the formula can work it out Conservation of Momentum mv = (m+M)v1 .028(230) = (.028 + 3.6) v1 v1 = 1.78 m/s Conservation of Energy KE = PE 1/2 (m+M)(v1)2 = (m+M)gh 1/2 (.028 + 3.6) (1.782) = (.028 + 3.6) (9.8) h h = .16 m L - h L = 2.8 L - h = 2.8 - .16 2.64 2 + x2 = 2.82 x = .93 m 93 cm L - h = 2.64 x

  3. v1 = 3.59(6.63) v1 = 3.59v1 m1v1 + m2v2 = (m1+ m2)v1 920v1+ 0= (1000 + 2300)v1 920v1 = 3300v1 v1 = 3.59v1 v1 = 23.79 m/s μ= 0.40 Wf = KE (This is v1) μmgx = 1/2 mv2 .8(9.8)(2.8) = 1/2 v2 v = 6.63 m/s 35. v1 ≈ 52mph m1v1 + m2v2 = 0 v1 = - 3 v2 38. Wf = KE Mass 1 μm1gx1 = 1/2 m1v12 Mass 2 μ(3)m1gx2 = 1/2 (3)m1v22 v1 = - 3 v2 x1 = v12 x1 v12 3x2 = 3v22 x2 = v22 = x2 v22 x1 x1 (-3)2 x1 (-3v2)2 = 9 = = x2 x2 x2 v22

  4. E1 m1v1x + m2v2x = (m1+ m2)vx 4.3(7.8) + 0= (4.3 + 5.6)vx 33.54= 9.9 v cosθ 3.39= v cosθ m1v1y + m2v2y = (m1+ m2)vy 0 + 5.6(10.2) = (4.3 + 5.6)vy 57.12 = 9.9 v sin θ 5.83= v sin θ 41. E1+E2 y direction v sin θ x direction v v1x θ v cos θ 5.83= v sin θ E2 v2y 3.39= v cosθ tan θ = 1.72 θ = 59.830 v2 = 3.392 + 5.832 v = 6.74 m/s v = 6.74 m/s @ 59.830 1.1 sin 300 y direction 42. mavay + mbvby = mavay1 + mbvby1 x direction mavax + mbvbx = mavax1 + mbvbx1 1.1 m/s a 1.8 m/s .4(1.8) + 0 = .4(1.1 cos 300) + .5 vbx1 0 = .4(1.1 sin 300) + .5 vby1 300 1.1 cos 300 a b vbx1 = .68 vby1 = -.44 .68 tan θ = .44/.68 θ = 32.900 .682 + .442 = v2 v = .81 m/s b θ .44 v = 0.81 m/s @ 327.10 v = 0.81 m/s 32.90 below the + x axis

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