A $100g$ granite cube slides down a $40°$ frictionless ramp. At the bottom, just as it exits onto a horizontal table, it collides with a $200g$ steel cube at rest. How high above the table should the granite cube be released to give the steel cube a speed of $150cm/s$?

Mass of granite,$m_1=100g=0.1Kg$
Mass of steel cube, $m_2=200g=0.2kg$
Speed of steel cube,${\left(v_{fx}\right)}_2=150cm/s=1.5m/s$

The angle of the slope,$\theta ={40}^{\circ }$

We need to figure out how far above the table the granite cube should move such that the steel cube collides with it at a specific velocity.

We will use both momentum and energy conservation to accomplish this. Consider a case in which a granite cube collides with a steel cube, and we treat it as an elastic collision. Object 2 is also at rest here. We don't need to think about the angle of the ramp because we need to think about the exact moment before and after the collision. As a result, we may compare this circumstance to a head-on collision.

Using the model's equation:

$\begin{array}{l}{\left(v_{fx}\right)}_2=\frac{2m_1}{m_1+m_2}{\left(v_{ix}\right)}_1\\ 1.5=\frac{2\times 0.1}{0.1+0.2}{\left(v_{ix}\right)}_1\\ {\left(v_{ix}\right)}_1=\frac{0.3\times 1.5}{0.2}=2.25m/s\end{array}$

Applying conservation principle,

$\begin{array}{l}{m}_{1}gh=\frac{1}{2}{m}_1{\left(v_{ix}\right)}_1{}^2\\ h=\frac{1}{2}\frac{{\left(v_{ix}\right)}_1{}^2}{g}\\ h=0.258m=25.8cm\end{array}$