Question

1 kg block slides down a frictionless inclined plane that makes an angle of 300 with respect to the ground. The total length of the plane is 2 m, but midway down it collides with a second block, weighing 0.5 kg. The two blocks stick together and travel as one unit the rest of the way down the ramp. What is the kinetic energy of the combined 1.5 kg block when it reaches the bottom of the plane

Answer 1

the kinetic energy of the combined 1.5 kg block when it reaches the bottom of the plane is 10.62 J

Step-by-step explanation:

Given that the data in the question;

angle of inclination with respect to the ground
= 30°

length of plane d = 2m

m₁ = 1 kg

m₂ = 0.5 kg

now, velocity of the first block at midpoint;

mv² = mgsin

v² = gsin

v² = gsin
d

v = √( gsin
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;

= (m₁ + m₂)gsin

+ Initial Kinetic energy

= (m₁ + m₂)gsin

+
mv₂²

we substitute

= [(1 + 0.5)9.8 × sin30 ×
] + [
× 1.5 × 2.0866 ]

= 7.35 + 3.2654

= 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