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;
[tex]P_{initial[/tex] = [tex]P_{final[/tex]
Now for BLOCK 1
mass = M₁ = M
KE = K[tex]_i[/tex]
[tex]\frac{1}{2}[/tex]mv[tex]_i[/tex]₁² = K[tex]_i[/tex]
we solve for v[tex]_i[/tex]₁
mv[tex]_i[/tex]₁² = 2K[tex]_i[/tex]
v[tex]_i[/tex]₁ = √( 2K[tex]_i[/tex] / m )
for BLOCK 2
mass = M₂ = 3m and since its at rest v[tex]_i[/tex]₂ = 0
Now after the collision; Total mass = m + 3m = 4m
KE = K[tex]_f[/tex]
[tex]\frac{1}{2}[/tex]( 4m )v[tex]_f[/tex]² = K[tex]_f[/tex]
(2m)v[tex]_f[/tex]² = K[tex]_f[/tex]
v[tex]_f[/tex] = √(K[tex]_f[/tex] / 2m)
so since [tex]P_{initial[/tex] = [tex]P_{final[/tex]
[m₁ × v[tex]_i[/tex]₁] + [m₂ × v[tex]_i[/tex]₂] = ( m + 3m ) × v[tex]_f[/tex]
so
[ m₁ × √( 2K[tex]_i[/tex] / m ) ] + [ m₂ × 0 ] = ( m + 3m ) × [ √(K[tex]_f[/tex] / 2m) ]
[ m × √( 2K[tex]_i[/tex] / m ) ] = 4m × [ √(K[tex]_f[/tex] / 2m) ]
square both side
m² × 2K[tex]_i[/tex] / m = (4m)² × K[tex]_f[/tex] / 2m
m² × 2K[tex]_i[/tex] / m = 16m² × K[tex]_f[/tex] / 2m
m × 2K[tex]_i[/tex] = 8m × K[tex]_f[/tex]
2K[tex]_i[/tex] = 8K[tex]_f[/tex]
K[tex]_f[/tex] = 2K[tex]_i[/tex] / 8
K[tex]_f[/tex] / K[tex]_i[/tex] = 2 / 8
K[tex]_f[/tex] / K[tex]_i[/tex] = 1 / 4
Therefore, the ratio Kf : Ki is 1 / 4 or 1 : 4
The ratio of the final kinetic energy to initial kinetic energy of the blocks is determined as 1:4.
The speed of the two blocks after collision is determined by applying the principle of conservation of linear momentum as follows;
m1u1 + m2u2 = v(m1 + m2)
mu1 + m2(0) = v(m + 3m)
mu1 = v(4m)
u1 = 4v
[tex]\frac{K.E_f}{K.E_i} = \frac{0.5(4m)v_f^2}{0.5(m)u_1^2} \\\\\frac{K.E_f}{K.E_i} = \frac{4(v_f)^2}{u_1^2} \\\\\frac{K.E_f}{K.E_i} = \frac{4v_f^2}{(4v_f)^2} = \frac{4v_f^2}{16v_f^2} = \frac{1}{4} = 1:4[/tex]
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