A block of mass m slides along a frictionless surface with kinetic energy Kiit collides with a block of mass 3m that is initially at rest, and the two blocks stick together and slide with total kinetic energy Kf . What is the ratio Kf:Ki

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

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