Answer:
the kinetic energy of the combined 1.5 kg block when it reaches the bottom of the plane is 10.62 J
Explanation:
Given that the data in the question;
angle of inclination with respect to the ground [tex]\theta[/tex] = 30°
length of plane d = 2m
m₁ = 1 kg
m₂ = 0.5 kg
now, velocity of the first block at midpoint;
[tex]\frac{1}{2}[/tex]mv² = mgsin[tex]\theta[/tex][tex]\frac{d}{2}[/tex]
[tex]\frac{1}{2}[/tex]v² = gsin[tex]\theta[/tex][tex]\frac{d}{2}[/tex]
v² = gsin[tex]\theta[/tex]d
v = √( gsin[tex]\theta[/tex]d)
g is 9.8 m/s
so we substitute
v = √( 9.8 × sin30° × 2)
v = √( 19.6 )
v = 3.13 m/s
Now, velocity just after collision of the blocks will be;
(m₁ + m₂)v₂ = m₁v
v₂ = m₁v / (m₁ + m₂)
we substitute
v₂ = (1 × 3.13) / (1 + 0.5)
v₂ = 3.13 / 1.5
v₂ = 2.0866 m/s
now, final kinetic energy will be;
[tex]KE_f[/tex] = (m₁ + m₂)gsin[tex]\theta[/tex][tex]\frac{d}{2}[/tex] + Initial Kinetic energy
[tex]KE_f[/tex] = (m₁ + m₂)gsin[tex]\theta[/tex][tex]\frac{d}{2}[/tex] + [tex]\frac{1}{2}[/tex]mv₂²
we substitute
[tex]KE_f[/tex] = [(1 + 0.5)9.8 × sin30 × [tex]\frac{2}{2}[/tex]] + [[tex]\frac{1}{2}[/tex] × 1.5 × 2.0866 ]
[tex]KE_f[/tex] = 7.35 + 3.2654
[tex]KE_f[/tex] = 10.62 J
Therefore, the kinetic energy of the combined 1.5 kg block when it reaches the bottom of the plane is 10.62 J