If the two blocks stick together after the collision, of the initial mechanical
ID: 2265507 • Letter: I
Question
If the two blocks stick together after the collision, of the initial mechanical energy of the (m1, m2 & spring) system: how much is lost to the environment during the collision?% (now fixed)how much is stored in the spring at maximum compression? % (now fixed) what is that maximum compression of the spring (when the blocks momentarily stop)? m If the two blocks instead collide elastically, what will be the speed and direction of each block immediately after the collision? block 1: m/s, of the initial mechanical energy of the system: how much is lost to the environment during the collision? % how much is stored in the spring at maximum compression? % (Where is the rest of the energy?) what will be the maximum compression of the spring afterwards? mExplanation / Answer
a)
by conservation of momentum.
m1v1 + m2v2 = m3v3 + m4v4
0+ 1.4*4 = 3.4*v
v = 1.647 m/s.
loss in kinetic energy =
= 1/2m2*v2*v2 -1/2 (m1+m2)*v*v
= 0.5*1.4*4*4 - 0.5*3.4*1.647*1.647
= 6.588 J.
= 6.588 / 11.2 = 58.82 %.
durind maximum compression velocity of both the blocks is zero.
the the complete kinetic energies of thr blocks is stored in the spring
which is 4.611 J.
4.611 / 11.2 = 41.169 %
let the maximum compression be X
then at maximum compression velocity is zero.
1/2(m1+m2)*v*v = 1/2*K*x*x.
0.5*3.4*1.647*1.647 = 0.5*130*x*x
x= 0.266 m.
b)
in elastic collosion energy is conserved.
m1V1 + m2v2 = m1v3 +m2v4
1.4*4 = 2*v3 + 1.4*v4.
1/2m2*v2*v2 = 1/2 m1v3*v3 + 1/2m2*v4*v4.
1.4*16 = 2*v3*v3 + 1.4*v4*v4.
v3 = 3.411 m/s---------------------velocity of block 1
velocity of block 2 = v4 = -0.874 m/s.
as energy is conserved loss in energy is zero.
at maximum compression velocity of block one is zero.
1/2 m1*v3*v3 = 1/2 k*x*.
0.5*2*3.411*3.411 = 0.5*130*x*x.
x = 0.432 m------------------maximum compression.
energy stored at maximum compression = 0.5*m1*v3*v3 = 0.5*2*3.411*3.411 = 11.634 J