# sol6 - Massachusetts Institute of Technology - Physics...

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Massachusetts Institute of Technology - Physics Department Physics-8.01 Fall 1999 Solutions for Assignment # 6 by Dru Renner [email protected] November 1, 1999 10:14 am Problem 6.1 (Ohanian, page 291, problem 13) The masses of the two automobiles are m 1 = 540 kg and m 2 = 1400 kg and their velocities, measured along the original direction of m 1 ,a re v 1 =80km / h 22 . 22 m / s and v 2 = 80 km / h ≈− 22 . 22 m / s. (a) We are told that the two automobiles remain locked together after the collision. After the collision both automobiles move as one, hence they both must have the same velocity which must also be the velocity of the center of mass ( v CM ). There are no outside forces, so the velocity of the center of mass remains unchanged during the collision. Therefore v = m 1 v 1 + m 2 v 2 m 1 + m 2 = 540 1400 540 + 1400 · 80 km / h 35 km / h 9 . 9m / s The velocity of the wreck is given by v . Note, it was unnecessary to convert from the non-metric unit “km / h” if we wanted the value of v in units of “km/h”. (b) The total kinetic energy before the collision is K before = m 1 v 2 1 2 + m 2 v 2 2 2 = 1 2 (540)(22 . 22) 2 + 1 2 (1400)(22 . 22) 2 4 . 8 × 10 5 J The total kinetic energy after the collision is K after = ( m 1 + m 2 ) v 2 2 8 . 2 × 10 4 J Note, it was necessary to convert from “km / h” to “m / s” to achieve an answer in units of “J”. (c) This calculation is best done in the center of mass frame. Since there are no outside forces, this frame is inertial; hence the acceleration in this frame is the same as the acceleration relative to the ground. We will make use of the kinematic relation a = v 2 v 2 0 2( x x 0 ) 1

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For the automobile of mass m 1 , its speed in the center of mass frame is v ± 1 = v 1 v CM =22 . 2 ( 9 . 9) = 32 . 1m / s The acceleration the passenger compartment experiences is given by a 1 = v ± 2 1 2 d 8 . 6 × 10 2 m / s 2 For the automobile of mass m 2 , its speed in the center of mass frame is v ± 2 = v 2 v = 22 . 2 ( 9 . 9) = 12 . 3m / s The acceleration the passenger compartment experiences is given by a 2 = v ± 2 2 2 d 1 . 3 × 10 2 m / s 2 Problem 6.2 (Ohanian, page 292, problem 23) This problem is illustrated in Figure 11.12 on page 292. Let m be the mass of each steel ball. The ±rst steel ball swings and gains kinetic energy until it collides with the second steel ball. Initially the ±rst steel ball is at rest at an angle θ relative to vertical, so its height relative to the bottom of the swing is h = l (1 cos θ ) Its mechanical energy is E = mgh = mgl (1 cos θ ) At the bottom of its swing, just before the collision, its mechanical energy is E = mv 2 2 Conservation of mechanical energy gives mgl (1 cos θ )= mv 2 2 = v = q 2 gl (1 cos θ ) Also in terms of just h , conservation of mechanical energy gives mgh = mv 2 2 = h = v 2 2 g (1) 2
(a) Because we are told the collision is elastic, we know that kinetic energy is conserved.

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## This note was uploaded on 11/20/2011 for the course PHY 203 taught by Professor Staff during the Fall '09 term at Rutgers.

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sol6 - Massachusetts Institute of Technology - Physics...

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