A 1.2 kg block is placed at rest on a ramp at some height above the bottom of th
ID: 1338189 • Letter: A
Question
A 1.2 kg block is placed at rest on a ramp at some height above the bottom of the track. It collides elastically with a 0.4 kg block at rest at the bottom of the track.(Figure 1) After colliding elastically both blocks make it safely around a loop-the-loop which has a radius 0.6 m. The track is frictionless.
What is the minimum height the 1.2 kg block must be released above the bottom of the track so that both blocks safely traverse the loop-the-loop?
Express your answer using three significant figures.
Explanation / Answer
total final energy of the blocks in closed system inital energy of block at rest = PE + KE
so mgh+0.5*m*(Vf^2- Vi^2)
final energy=potential energy + kinetic energy of bodies
= 0.5* (1.2+0.4) v^2
to complete circle it should have velociy
v=sqrt(5*r*g) = (5*0.6*9.8)^0.5
V =5.42 m/s
apply the law of Conservation of energy
as total initial energy = final energy
then 1.2*9.8*h + 0.5 *1.2*(5.42)^2 =0.5*(1.6)*(5.42)^2
thus solving this for h we get h=0.4995 -----------<<<<<<<<<<Answer
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