Tutorial 5 Solutions

Tutorial 5 Solutions - PHYSICS 1010 T#5 Questions and...

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PHYSICS 1010 T#5 Questions and Solutions Chapter 9, Problem 63 A body of mass 2.0 kg makes an elastic collision with another body at rest and continues to move in the original direction but with 1/4 of its original speed. (a) What is the mass of the other body? (b) What is the speed of the two-body center of mass if the initial speed of the 2.0 kg body was 4 m/s? 63. (a) Let m 1 be the mass of the body that is originally moving, v 1 i be its velocity before the collision, and v 1 f be its velocity after the collision. Let m 2 be the mass of the body that is originally at rest and v 2 f be its velocity after the collision. Then, according to Eq. 9-67, We solve for m 2 to obtain We combine this with to obtain . (b) The speed of the center of mass is . Chapter 9, Problem 65 In Figure particle 1 of mass m 1 = 3.0 kg slides rightward along an x axis on a frictionless floor with a speed of 2.0 m/s.When it reaches x = 0, it undergoes a one-dimensional elastic collision with stationary particle 2 of mass m 2 = 0.4 kg. When particle 2 then reaches a wall at x w = 70 cm, it bounces from the wall with no loss of speed. At what position (in cm) on the x -axis does particle 2 then collide with particle 1? 65. We use Eq 9-67 and 9-68 to find the velocities of the particles after their first collision (at x = 0 and t = 0): v 1 f = v 1 i = (2.0 m/s) = m/s v 2 f = v 1 i = (2.0 m/s) = m/s 1.7 m/s. At a rate of motion of 1.7 m/s, 2 x w = 140 cm (the distance to the wall and back to x = 0) will be traversed by particle 2 in 0.82 s. At t = 0.82 s, particle 1 is located at x = (–2/7)(0.82) = –23 cm,
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and particle 2 is “gaining” at a rate of (10/7) m/s leftward; this is their relative velocity at that time. Thus, this “gap” of 23 cm between them will be closed after an additional time of (0.23 m)/(10/7 m/s) = 0.16 s has passed. At this time ( t = 0.82 + 0.16 = 0.98 s) the two particles are at x = (–2/7)(0.98) = –28 cm. CHAPTER 10 ROTATION Linear position: x Angular position: θ Linear displacement: Δx Angular displacement: Δθ Linear speed: v Angular speed: ω Linear acceleration: a Angular acceleration: α Mass: m – is a measure of the tendency of an object to resist changes in its linear motion. Moment of inertia: I – is a measure of the resistance of an object to changes in its rotational motion. I = Σ m i r i 2 Force: F Torque: τ – measures the tendency of a force to rotate an object about an axis. τ = rFsin(,) Linear momentum: p Angular momentum: L The net external force on a system of particles equals the time rate of change of the total linear momentum of the system: ΣF ext = dp total /dt The net external torque acting on a system about some axis passing through an origin in an inertial frame equals the time rate of change of the total angular momentum of the system about that origin Στ ext = dL total /dt Law of conservation of linear momentum : the total linear momentum of a system of particles is conserved if no net external force is acting on the
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This note was uploaded on 10/02/2011 for the course PHY 1010 taught by Professor Rupinder during the Spring '09 term at UOIT.

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Tutorial 5 Solutions - PHYSICS 1010 T#5 Questions and...

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