Exam 3 Solutions

# Exam 3 Solutions - midterm 03 ALIBHAI ZAHID Due Apr 4 2007...

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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 . 23 . 4 kg 1 . 8 m 3 . 6 m θ 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 = m g = 229 . 32 N , and μ = 0 . 35 .

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midterm 03 – ALIBHAI, ZAHID – Due: Apr 4 2007, 11:00 pm 2 Pivot N w f N f m g 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 μ W W d L tan θ . Thus, solving for θ , we have θ arctan bracketleftbigg d μ L 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 . A solid cube is resting on a horizontal sur- face. The cube is constrained to rotate about an axis at its bottom left edge (due to a small obstacle on the table). A bullet with speed v min is shot at the left-hand face at a height of 4 3 a . The bullet gets embedded in the cube.
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