{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

S08 AM261 MT S - Phys 263/Amath 261 SOLUTIONS Midterm 1(a...

Info icon This preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon
Phys 263/Amath 261 Midterm SOLUTIONS 1. (a) To find the CM coordinates use R = 1 M k m k r k (1) = 1 m 1 + m 2 ( m 1 r 1 + m 2 r 2 ) (2) = 1 m 1 + m 2 2 t 2 m 1 + (1 + t 2 ) m 2 , 3 tm 1 , 4 m 1 + 4 t 2 m 2 (3) (b) V = ˙ R = 1 m 1 + m 2 [4 tm 1 + 2 tm 2 , 3 m 1 , 8 tm 2 ] (4) and A = ˙ V = 1 m 1 + m 2 [4 m 1 + 2 m 2 , 0 , 8 m 2 ] (5) (c) The linear momentum is P = M V = [8 , 3 , 16] (6) 2. To calculate the velocity and position requires two integrations, and solving of the two integration constants via initial conditions. (a) To get the velocity e - λt = m ˙ v = m dv dt (7) dv dt = 1 m e - λt (8) v ( t ) = 1 m - 1 λ e - λt t 0 (9) v ( t ) = - 1 ( e - λt - 1 ) (10) (b) Integrate again to get the position v = dx dt (11) x ( t ) = - 1 - 1 λ e - λt - t t 0 (12) x ( t ) = 1 1 λ e λt + t - 1 λ (13) 1
Image of page 1

Info icon This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
(c) As t approaches infinity: v ( t → ∞ ) 1 ( e -∞ + 1 ) = 1 (14) (d) The velocity we are looking for is v ( t ) = 1 - e - 1 (15) 1 - e t/ 2 = 1 - e - 1 (16) which implies t = 2 (17) 3. (a) The potential is U ( x ) = k 2 x 2 - k 4 a 2 x 4 (18) (b) The equilibrium positions are found by solving dU dx = 0 (19) kx - k a 2 x 3 = 0 (20) x ( x 2 - a 2 ) = 0 (21) giving equilibrium points at x = 0 and x = ± a (22)
Image of page 2
Image of page 3
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}