Answer:
the ratio Kf : Ki is 1 / 4 or 1 : 4
Explanation:
Given the data in the question;
Since this is a perfectly inelastic collision, momentum is conserved;
=
Now for BLOCK 1
mass = M₁ = M
KE = K
mv₁² = K
we solve for v₁
mv₁² = 2K
v₁ = √( 2K / m )
for BLOCK 2
mass = M₂ = 3m and since its at rest v₂ = 0
Now after the collision; Total mass = m + 3m = 4m
KE = K
( 4m )v² = K
(2m)v² = K
v = √(K / 2m)
so since =
[m₁ × v₁] + [m₂ × v₂] = ( m + 3m ) × v
so
[ m₁ × √( 2K / m ) ] + [ m₂ × 0 ] = ( m + 3m ) × [ √(K / 2m) ]
[ m × √( 2K / m ) ] = 4m × [ √(K / 2m) ]
square both side
m² × 2K / m = (4m)² × K / 2m
m² × 2K / m = 16m² × K / 2m
m × 2K = 8m × K
2K = 8K
K = 2K / 8
K / K = 2 / 8
K / K = 1 / 4
Therefore, the ratio Kf : Ki is 1 / 4 or 1 : 4