A penguin pushes with its feet to gain an initial speed of 1.0 m/s and slides for 0.5 m across the flat surface before stopping. If the same penguin on the same surface next slides for 2.0 m before stopping, what was its initial speed this time? If you double the speed, you have four a) 0.5 m/s times as much energy to slide. b) 1.0 m/s 1 E1 = - mv2 c) 2.0 m/s 2 d) 3.0 m/s Let v2 = 2v1 1 e) 4.0 m/s E2 = - mus 2 1 E2 = - m (2v1 )2 = -m402 = 4 mv? 2 N E2 = 4E1Conservation of momentum in two dimensions - example Two billiard balls with the same mass m collide. The cue ball has an initial speed of 1.0 m/s. The 8 ball is at rest. If the cue ball moves off at 53 to its initial direction at speed |v, | and the 8 ball moves off at 33 to the cue ball's initial direction at speed | 12 ], what are their speeds (|v, | and | v2)) immediately after the collision? +x 1.0 m/ before after Conservation of momentum in two dimensions - example ty x -component: miVixo + mav2xo = miVix + mzV2x + * V1x0 - Vix = V2x 32 V1x0 - |14 | cos 530 = |72| cos 33 V1x - v1| cos 53 before after cos 330 11x0 21 / COS 530 Ivy | sin 53 V1x0 | | cos 53 cos 33 Cos 330 sin 33 Cos 33 cos 33- 1 y-component: 1.466/D1/ = 0.838 -0.717| vil miViyo + mavzyo = milly - mavzy 2.183/v1| = 1.192 |71| = 0.55 m/5 Vzy = Vly | v2 | sin 330 = |v1 | sin 530 (0.55 m/s) sin 530 v1 sin 53 1021 = sin 330 sin 33 |v2| = 0.80 m/sConservation of momentum - example ty x -component: (m1 + m2 + m3)Vox = miVix + m2Vzx + m3V3x + X 14 m/s 0 = V2x + V3x V3x = -V2x = -(24 cos 220) 1 V3x = -22.25 m/s 3 220 2 y-component: 24 m/s (m1 + m2 + m3)voy = mivly - M2 zy + m3V3y 0 = Vly - Vay + 13y V3y = Vzy - Vly = (24 sin 220) - 14 A V3y V3y 1 V3y = -5.009 m/s V3 1031 = (V3x) 2 + (V3y ) = 23 m/s 0 = tan-1([v3yl/|v3x1) = 13 0 = 13 [below the negative x - axis]