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sample_exam_hrly_I_solution

# sample_exam_hrly_I_solution - Assume the block B slides...

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TAM212 Introductory Dynamics Spring 2005 S. Balachandar Sample Exam (Hourly I) A1 a n = v 2 ρ = 4 2 2 = 8 m / s 2 F n = - ma n ˆ e r N = - 10(8) ˆ e r N = - 80 ˆ e r N F n = - 80 ˆ e r N A2 W = T mgh = 1 2 mv 2 f - 1 2 mv 2 i 10(9 . 81)(2) = 1 2 10( v 2 f - 4 2 ) 39 . 24 = v 2 f - 16 v f = 7 . 43 m/s A3 W = Force · Distance = T cos 60(10) = 5 T N-m 1

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A4 There is no acceleration in the x direction and so the x component of the velocity remains con- stant. So the ball lands in the cart with v x = v o m/s. Using conservation of momentum ( m + m ) v x f , bc = mv xi , b + mv xi , c 2 v x f , bc = v 0 + 0 v x f , bc = v 0 2 m / s A5 y = x 3 dy dt = d ( x 3 ) dt ˙ y = 3 x 2 ˙ x ˙ y | 2 = 3(2) 2 5 = 60 cm/s 2
B1 First find the velocity with which block A hits block B W = T mgH = 1 2 m ( v 2 f - v 2 i ) (9 . 81)6 = 1 2 ( v 2 f - 0) v f = 10 . 84988 m/s Coefficient of restitution e = v B f - v A f v Bi - v Ai 0 . 6 = v B f - v A f 10 . 84988 - 0 v B f - v A f = 6 . 50993 Conservation of momentum m A v Ai + m B v Bi = m A v A f + m B v B f m (0) + m (10 . 84988) = m ( v A f + v B f ) v A f + v B f = 10 . 84988 Solving the two equations v B f = 8 . 67991 m/s v A f = 2 . 16998 m/s This is the speed with which block B starts sliding.

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Unformatted text preview: Assume the block B slides distance x be±ore coming to rest. Using conservation o± energy W f riction = ∆ T-F f riction x = 1 2 m ( v 2 f-v 2 i )-μ Nx = 1 2 m (0-v 2 i )-. 2(12)(9 . 81) x =-1 2 (12)(8 . 67991) 2 x = 19 . 2 m 3 B1 θ = t 2 ˙ θ = 2 t ¨ θ = 2 r = 4 + θ = 4 + t 2 ˙ r = 2 t ¨ r = 2 When θ = π/ 3, t = p π/ 3 = 1 . 02333 v p = ˙ r ˆ e r + r ˙ θ ˆ e θ = 2 t ˆ e r + (4 + t 2 )2 t ˆ e θ = 2 t ˆ e r + (8 t + 2 t 3 ) ˆ e θ v p ± ± ± θ = π/ 3 = 2(1 . 02333) ˆ e r + (8(1 . 02333) + 2(1 . 02333) 3 ) ˆ e θ = 2 . 04665 ˆ e r + 10 . 32986 ˆ e θ 4...
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sample_exam_hrly_I_solution - Assume the block B slides...

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