All three parts please MasteringPhysics: Ch 07 HW (Part 1) - Google Chrome https

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All three parts please

MasteringPhysics: Ch 07 HW (Part 1) - Google Chrome https://session.masteringphysics.com/myct/itemView?assignmentProblemD-51476055 A COLLEGE PHYS-203 fa2015 Ch 07 HW (Part 1 Problem 7.79 Signed in as Isaac Lamberth Help Close Resources 14 of 21 next » (C Problem 7.79 Part A Determine the speed of the block with mass m = 1.90 kg after the collision A block of mass m = 1.90 kg slides down a 30.0° incline which is 3.60 m high. At the bottom, it strikes a block of mass M 6.50 kg which is at rest on a horizontal surface (Figure 1) (Assume a smooth transition at the bottom of the incline.) The collision is elastic, and friction can be ignored Express your answer to three significant figures and include the appropriate units. IA um= / Value Units Submit My Answers Give Up Part B Determine the speed of the block with mass M = 6.50 kg after the collision Express your answer to three significant figures and include the appropriate units UM= Value | Units | Figure 1 ofl Submit My Answers Give Up Part C Determine how far back up the incline the smaller mass will go Express your answer to three significant figures and include the appropriate units. 3.60 m 30.0 dA= Value Units Submit My Answers Give Up 8:49 PM 11/3/2015 Search the web and Windows

Explanation / Answer

final Velocity of the both the blocks - Vf^2 = 2gH cos 30

Vf^2 = 2* 9.81 * 3.6 * cos 30

Vf = 7.82 m/s

so velocity of smaller mass = velocity of larger mass = 8.404 m/s

Momentum .initial = 7.82 * 1.9 = 14.878 kg*m/s

total INitail Momentum = P1 + 0 = 14.878

P.inital=P.final (elastic collision)

1.9 V1 + 6.5 * V.2 = 14.878

V1 = 7.83 - 3.42 V2 ------------------------1

Using Conservation of Energy

0.5 *1.9 * 7.82^2 = 0.5 *1.9 * v1^2 + 0.5 * 6.5 * (V2)^2

58.1 = 0.95 * (7.83 - 3.42V)^2 + 3.25 V^2

V2 spped of larger mass = 3.53 m/s

V1 = 7.83 - (3.42* 3.53) = 4.24 m/s

--------------------

distance travelled by mass1 = R = 3.6/sin 30 = 7.2 m

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