Inelastic collisions, conservation of energy? Starting from rest, a block of mas
ID: 3893107 • Letter: I
Question
Inelastic collisions, conservation of energy?
Starting from rest, a block of mass m slides without friction down a quarter-circle ramp of radius r, as shown at right. At the bottom of the ramp it strikes a second block of mass 2m, initially perched at rest a distance R above the floor. If the collision between the two blocks is perfectly elastic, i.e., if the total kinetic energy is conserved in the collision, (a) Find the distance l a from the base of the cliff where the mass 2m strikes the floor, (b) Find the distance l b where the mass m ultimately strikes the floor, (c) If instead the collision is perfectly inelastic, i.e., if m an d 2m stick together up on colliding, find the distance l c where they together strike the floor.Explanation / Answer
Potential energy = Kinetic energy
mgr=mv^2/2
so for the first object v=sqrt(2gr)
a) elastic collision
mv=2mu
velocity for second object is u=v/2
time of flight in x direction
t = l_a/u
time of flight in y direction is from
R=qt^2/s
t = sqrt(2R/g)
t=t
l_a=u*sqrt(2R/g)=1/2*sqrt(2gr)*sqrt(2R/g)=sqrt(r*R)
b)when we have elastic collision, than mass m will stop on top of cliff
l_b=0
c) inelastic collision
mv=(2m+m)u
u=v/3
l_c = 2*sqrt(r*R/3)