A 2.2 kg block that starts at rest is 1.2 m above the first horizontal part of 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.

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.