A 2.2 kg block that starts at rest is 1.2 m above the first horizontal part of a
ID: 1427198 • Letter: A
Question
A 2.2 kg block that starts at rest is 1.2 m above the first horizontal part of a frictionless track. Once block 1 reaches that first horizontal part, it collides with a second block of mass 3.1 kg that is initially at rest. Block 1 proceeds to slide to a max height of 0.0343 m back up the first ramp while block 2 is sent down a second incline that is 2.1 m above the 5 m long base of the track. Once it reaches the base, block 2 experiences friction. The coefficient of kinetic friction between the base of the track and block 2 is 0.728. At the end of the 5 m long base of the track is a cliff. Does block 2 stop in time, or does it go over the cliff?
Explanation / Answer
Using Coservation of energy,
Speed of Block 1 before collision = sqrt ( 2 g h1 ) = 4.85 m/s where h1=1.2 m
Speed of Block 1 after collision = sqrt ( 2 g h2 )= 0.82 m/s in reverse direction
Change in Momentum for block 1 = m(v2-v1) = 12.474 Kg m/s
By conservation of momentum , this would be change in momentum for the second block.
Its initial velocity = 0
Its veloctiy after collision = 12.474/m2 =4.02 m/s
By conservation of energy, its velocity after sliding down the second incline = sqrt ( 0.5* 4.02^2 + gh) =14.63 m/s
Frictional force now would be uN = 0.728*m2g =22.11 N
Work Done by friction to move 5 m = f*5m = 110.58 J
Kinetic energy of block befor friction = 0.5mv^2 = 331.75 J > Work done by fricition
Thererfore it will go on the cliff. Since we dont know the height of cliff we cant se whether it will reach the top or not.