Question

A 0.100 kg limestone cube is released from rest, and proceeds to slide down a frictionless ramp. At the bottom of the ramp, the limestone cube makes an elastic collision with a steel cube whose mass is 0.200 kg, which is initially at rest. At what vertical height should the limestone cube be placed such that the steel cube has a velocity of 1.50 m/s after the collision

Answer 1

Answer:

the height at which the limestone cube must be placed is 0.23 m.

Explanation:

Given;

mass of limestone cube, m₁ = 0.1 kg

initial velocity of the limestone cube, u₁ = 0

mass of steel cube, m₂ = 0.2 kg

initial velocity of the steel cube, u₂ = 0

final velocity of the steel cube, v₂ = 1.5 m/s

Apply the principle of conservation of energy to determine the height of the limestone cube.

Potential energy of the limestone cube at top = Kinetic energy of steel cube at base

m₁gh = ¹/₂m₂v₂²

where;

h is the height at which the limestone cube is placed

$h = \frac{m_2v_2^2}{2m_1g} \\\\h = \frac{0.2\times 1.5 ^2}{2 \times 0.1 \times 9.8} \\\\h =0.23 \ m$

Therefore, the height at which the limestone cube must be placed is 0.23 m.