AU17 PHYSICS 1250-N × . HW B Serway, physics for Scie ,\' M light spring of forc

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AU17 PHYSICS 1250-N × . HW B Serway, physics for Scie ,' M light spring of force c C A Light Spring Of Force x two blocks connected b rlin.png (599*509 x . www.webassion.netwe.. blecpp That Disolays A Tab (D) What is the height h of the wedge? Need Help?Read ItWatch t 9. 2/6 paints PrEviaus Answers SePSE9 9 P.047 My N A light spring of force constant 3.95 N/m is compressed by 8.00 cm and held between a 0.250 kg block on the left and a 0.510 kg block on the right. Both blocks are at rest on a horizontal surface. The blocks are released simutaneously so that the spring tends to push them apart. Find the maximum velocity each block attains if the coefficient of kinetic friction between each block and the surface is the following. In each case, assume that the coefficient of static friction is greater than the coefficient of kinetic friction. Let the positive direction point to the right. 0.250 kg block 0.510 kg block 0.000 -2605 0.100 0.456 m/s 128 mys mys m/'s mys mys Need Help? Submit Answer Save Progress Practice Another Version 10, -2 points SarPSE89,P136, Mty N o automobiles of equal mass approach an intersection. One vehicle is traveling with velocity 13.0 m/s toward the east and the other is traveling north with speed v . Neither driver sees the other. The vehicles colli de in the intersection and stick together, leaving parallel skid marks at an angle of 59.5 narth of east. Determine the initial speed z of the northward-maving vehicle. m/s The speed limit for both roads is 35 mi/h and the driver of the northward-moving vehicle claims he was within the speed limit when the collision occurred. Is he telling the truth? Yes No Need Help?Read It 6:45 PM Type here to search 1/16/2017

Explanation / Answer

10.

consider the east-west direction along X-axis and north-south direction along Y-axis

using conservation of momentum along the Y-direction

m (0) + m (v2) = (2m) V Sin59.5

v2 = 2 V Sin59.5 eq-1

along the X-direction

m (13) + m (0) = (2m) V Cos59.5

13 = V (2 Cos59.5)

V = 12.8 m/s

using eq-1

v2 = 2 V Sin59.5 = 2 (12.8) Sin59.5 = 22.1 m/s

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