Exam 3 Solutions - midterm 03 – ALIBHAI ZAHID – Due Apr...

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Unformatted text preview: midterm 03 – ALIBHAI, ZAHID – Due: Apr 4 2007, 11:00 pm 1 Question 1 part 1 of 1 10 points Consider the collision of two identical par- ticles, where the initial velocity of particle 1 is v 1 and particle 2 is initially at rest. 1 2 v 1 After an elastic head-on collision, the final velocity v ′ 2 of particle 2 is given by 1. v ′ 2 = 5 v 1 3 2. v ′ 2 = 2 v 1 3. v ′ 2 = 3 v 1 4 4. v ′ 2 = 4 v 1 3 5. v ′ 2 = v 1 4 6. v ′ 2 = v 1 3 7. v ′ 2 = v 1 2 8. v ′ 2 = v 1 correct 9. v ′ 2 = 0 10. v ′ 2 = 2 v 1 3 Explanation: For the final velocity v ′ 2 of particle 2 after an elastic collision, we have v ′ 2 = 2 v cm − v 2 . For the present case, v cm = m 1 v 1 + m 2 v 2 m 1 + m 2 = v 1 2 . So v ′ 2 = 2 parenleftBig v 1 2 parenrightBig − 0 = v 1 . Question 2 part 1 of 1 10 points A 23 . 4 kg person climbs up a uniform ladder with negligible mass. The upper end of the ladder rests on a frictionless wall. The bottom of the ladder rests on a floor with a rough surface where the coefficient of static friction is 0 . 35 . The angle between the horizontal and the ladder is θ . The person wants to climb up the ladder a distance of 1 . 8 m along the ladder from the ladder’s foot. The acceleration of gravity is 9 . 8 m / s 2 . b 23 . 4 kg 1 . 8 m 3 . 6 m θ b μ = 0 . 35 μ =0 Note: Figure is not to scale. What is the minimum angle θ min (between the horizontal and the ladder) so that the person can reach a distance of 1 . 8 m without having the ladder slip? 1. 51 . 6692 ◦ 2. 53 . 2948 ◦ 3. 55 . 008 ◦ correct 4. 56 . 7466 ◦ 5. 58 . 5729 ◦ 6. 60 . 389 ◦ 7. 62 . 2685 ◦ 8. 64 . 3008 ◦ 9. 66 . 3636 ◦ 10. 68 . 5032 ◦ Explanation: Let : d = 1 . 8 m , L = 3 . 6 m , W = mg = 229 . 32 N , and μ = 0 . 35 . midterm 03 – ALIBHAI, ZAHID – Due: Apr 4 2007, 11:00 pm 2 b Pivot b N w f N f mg θ Using the above equilibrium conditions, we have (taking the sum of the torques at the bottom of the ladder) summationdisplay F x : f − N w = 0 , (1) summationdisplay F y : N f − W = 0 , and (2) summationdisplay τ ◦ : W d cos θ − N w L sin θ = 0 , (3) where d is the distance of the person from the bottom of the ladder. Using Eq. 2, Eq. 3 becomes f L sin θ = W d cos θ , so f = W d L tan θ . (4) The ladder may slip when f = f max = N w , from Eq. 4 f max ≡ μW ≥ W d L tan θ . Thus, solving for θ , we have θ ≥ arctan bracketleftbigg d μL bracketrightbigg (5) ≥ arctan bracketleftbigg (1 . 8 m) (0 . 35) (3 . 6 m) bracketrightbigg ≥ 55 . 008 ◦ . Question 3 part 1 of 1 10 points Assume: A bullet of mass m and cube of mass M undergo an inelastic collision, where m ≪ M . Note: The moment of inertia of this cube (with edges of length 2 a and mass M ) about an axis along one of its edges is 8 M a 2 3 ....
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This note was uploaded on 10/13/2009 for the course PHY 303K taught by Professor Turner during the Spring '08 term at University of Texas.

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Exam 3 Solutions - midterm 03 – ALIBHAI ZAHID – Due Apr...

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