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Elastic Collision Analysis in Physics

Solve elastic collision equations to calculate velocity and kinetic energy for two masses colliding. Conservation of energy and momentum principles are applied for detailed analysis. Examples and formulas provided for comprehensive understanding.

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Elastic Collision Analysis in Physics

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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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