Download Angular Kinematic Equations - Fundamental of Physics - Solved Past Paper and more Exams Physics Fundamentals in PDF only on Docsity! Department of Physics University of Maryland College Park, Maryland PHYSICS 121 Prof. S. J. Gates Fall 2002 Exam II Solutions Nov. 11, 2002 Section I. Multiple Choice Questions Each question in this section is worth eight (8) points. You should NOT take more than two minutes per question. If you do, it is advisable to continue on to the next question! (1.) (a.) Since a = v2/R and v = 2 π R T implies a = 4 π2 R T 2 , we find a = 4 π2× 4m (4 s)2 = π2 ms2 = 9.87 m s2 . (2.) (a.) As with the problem on the practice exam, the velocities and heights are related by (V1) 2 = 2 g H1 and (V2) 2 = 2 g H2 so that H1 H2 = (V1) 2 (V2)2 = [(2 ms ) 2] / [(6 ms ) 2] = 1/9. (3.) (b.) Since both start from rest W1 = 1 2 m1 (V1) 2 and W2 = 1 2 m2 (V2) 2 and since W1 = W2 it follows that 3 kg (V1) 2 = 27 kg (V2) 2 or V1 = 3 V2. (4.) (b.) Here we have the data; ω0 = 2π rad s , ωf = 4π rad s , θ0 = 0 and θf = 6π rad and one of the basic angular kinematic equations is (ωf ) 2 = (ω0) 2 + 2 α (∆θ). Putting in the numbers yields α = π rad s2 , also since ω = ω0 + α t = 2π rad s + ( π rad s2 ) t we see that the value of t = 2 s works. (5.) (d.) We can find the work done by using W = ∆KE = 12 m [0 − (v0) 2] = − 12 (2000 kg) (6 m s ) 2 = −36000 J . The time over which the col- lision occurs is ∆t = 0.12 s. So the power delivered is ∣∣∣W ∣∣∣/∆t = 300,000 J . Section II. Analytical Questions Problem (1.) From Newton’s second law it follows that m a = F and the problem gives the force and acceleration. Therefore, it must be so that m (9 ms4 t 2) = 36 (Ns3 t 3) → m = 4 t kgs and we see that the mass depends on the time t. The problem also states that the bucket can only hold 16 kg when full. So we have 16 kg = 4 t kg s → t = 4 s Problem (2.) This is a momentum conservation problem. The momenta for ball #1 and ball #2 before the collision are given by Momentum x-comp. y-comp. ~p1 10 kg (7 m s ) 0 ~p2 0 5 kg (14 m s ) ~PBeforetot 70 ( kg m s ) 70 ( kg m s ) After the collision the two balls are stuck together so that Mt = 15 kg but we don’t know the x-component of velocity V ′x nor the y-component of veloc- ity V ′y Momentum x-comp. y-comp. ~PAftertot 15 kg V ′ x 15 kg V ′ y Momentum conservation implies that 15 kg V ′x = 70 ( kg m s ) , 15 kg V ′ y = 70 ( kg m s ) , V ′x = 14 3 ( ms ) , V ′ y = 14 3 ( ms ) , and to find the angle we note tanθ = V ′y V ′x = (14/3) (14/3) = 1 → θ = 45o. Problem (3.) 2
