Mastering Physics: Ch 07 X o Session ring physics.com /myct/itemView? assignment

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Mastering Physics: Ch 07 X o Session ring physics.com /myct/itemView? assignment Problem 51476080 COI I FGF PHYS-203 fa2015 Mastering Physics: Ch 07 X o Session ring physics.com /myct/itemView? assignment Problem 51476089 A COLLEGE PHYS-203 fa2015 Ch 07 HW (Part 2 Problem 7.44 Problem 7.44 Correct Billiard ball A of mass m A 0.125 kg moving with speed vA 2.80 m/s strikes ball B, initially at rest, of mass mB 0.136 kg. As a result of the collision, ball A is deflected off at an angle of Part C 30.0° with a speed va 2.10 m/s, and ball B moves with a speed vR at an angle of da to original direction of motion of ball A. Solve these equations for the angle, da, of ball Bafter the collision. Do not assume the collision is elastic. Express your answer to three significant figures and include the appropriate units. Value Units Submit My Answers Give Up Part D Solve these equations for the speed, vB, of ball Bafter the collision. Do not assume the collision is elastic. Express your answer to three significant figures and include the appropriate units. Value Units Sianed in as Chris Lowther I Helo l Close Signed in as Chris Lowther I Help I Close Resources previous l 8 of 19 l next

Explanation / Answer

Conservation of Momentum

Assume Va = 2.8 m/s +x

Initial momentum = 0.125 * 2.8 + 0.136 * 0 = 0.35 kg m/s +x

Final momentum = 0.35 Ns +x

momentum in x = 0.125 * 2.1 * cos 30 + 0.136 * Vbx = 0.36 Ns

Vbx = 0.9755 m/s

momentum in y = 0.125 * 2.1 * sin 30 + 0.136 * Vby = 0

Vby = -0.965 m/s

theta = tan^-1(Vby/Vbx) = 44.69 degree = 45.0 degree (below the x axis )

part b )

V = sqrt(Vbx^2 + Vby^2)

V = 1.37 m/s

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