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2009_3rd_exam_PH-211_SOLUTION

# 2009_3rd_exam_PH-211_SOLUTION - Exam#3 Monday July 13 2009...

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Exam #3 PHYSICS 211 Monday July 13 th , 2009 NAME ____________________________ Please write down your name also on the back page of this exam 1. 1A The figure shows a three particle system with masses m 1 = 8.0 kg, m 2 = 4 kg, and m 3 = 8 kg. What are the coordinates of the center of mass? 2 X (m) 3 - 1 1 0 1 2 3 4 5 6 7 Y(m) m 2 = 4 . 0 kg m 3 = 8 . 0 kg m 1 = 8 . 0 kg ANSWER: R CM = 1 m + 1 m i j 1B Find the coordinates of the "Center of Mass" of the homogeneous thin plate shown in the figures. 6 cm 12 cm i j ANSWER: R CM = - 1 cm + 1 cm i j O ________________________________________________________________________

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2. 2A Two bodies have undergone an elastic one-dimensional collision along the x -axis. The figure shows the position vs time for those bodies and for the center of mass. P x t S U Q R T Which segments correspond to the motion of the center of mass ANSWER ___ S and R ______ If we assume that the mass of the particles are equal, indicate the segments corresponding to the motion of the particle that was moving faster before the collision. ANSWER __ U and T _______ 2B The figure below shows groups of identical particles that move parallel to either the x-axis or the y-axis and at identical speeds. Rank the groups according to the speed of the center of mass, greatest first . P S Y X Y X Y X Y X U T ANSWER __ T,U,P,S _______
3. Given the vectors A = 1 i + 2 j + 1 k and B = 2 i - 1 j - 1 k 3A . Calculate and graph the vectors C and D , where C = ( B + A ) ( 3 B A ) and D =(4 A + B ). ( 2 A B ) Z Y X C = -2B +2A = - 2(B-A) D= 2A + 2B = 2( A + B) B -A = i -3 j -2k C = - 2(B – A) = - 2i + 6j + 4 k B+A = 3i + j + 0 D = 2(A+B) = 6i + 2j + 0 k D C 1 2 1 2 2 1 3B . Calculate and graph the vector D X C, where X stands for the vector product Z Y X = 8 i -24j + 40 k DxC = - 2 6 4 i j k 6 2 0 8 16 8 16 16 8 DxC 8D 8C

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4. 4A A ball of mass m = 0.1 Kg approaches a wall with a speed of 2 m/s. The incidence direction makes 30 degrees with the vertical direction as shown in the figure. After bouncing from the wall the ball also has a speed of 2 m/s. The change of the LINEAR MOMENTUM of the particle is: ANSWER: P = - 0.2 Kg m/s + 0 j i 2 m/s m = 0.1 Kg 2 m/s 30 30 i j 4B The figure shows three particles of the same mass and the same constant speed. The arrows I the figure indicate the corresponding velocity vectors.
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2009_3rd_exam_PH-211_SOLUTION - Exam#3 Monday July 13 2009...

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