In a physics lab, a small cude slides down a frictionless incline, and elastically strikes a cube at the bottom of the inclince that is only one-half its mass. If the incline is 30 cm high and the table is 90 cm off the floor, where does each cube land?
I got the V0 to be 2.43 from mhg = 1/2mv^2 but after setting up M1V1=M1V1' + M2V2' i don't know what to do
The speed of the sliding cube at impact is V = sqrt(2gH) = 2.42 m/s
Next you have to figure out the velocities of the sliding cube (mass M) and the impacted cube (Mass M/2), after impact. You need to write equations of both momentum and kinetic energy conservation. You will find, if you do it right, that the smaller cube travels at 4V/3 and the larger cube at V/3.
Finally, use those horizontal velocities and the height of the table to predict where they land.
To solve this problem, you correctly started by finding the initial velocity (V0) of the first cube using the conservation of mechanical energy equation, which states that potential energy (mgh) is equal to kinetic energy (1/2mv^2).
Now, let's proceed to the next step using the principle of conservation of linear momentum. This principle states that the total momentum of an isolated system remains constant before and after a collision.
In this case, the collision is an elastic collision because it is stated that the cubes collide elastically. During an elastic collision, both momentum and kinetic energy are conserved.
Let's denote the mass of the first cube as M1 and the mass of the second cube as M2. We know that M2 is half the mass of M1.
Initially, the first cube (M1) has a velocity V0, and the second cube (M2) is at rest.
Using the conservation of linear momentum:
M1 * V0 = M1 * V1' + M2 * V2'
Here, V1' and V2' represent the final velocities of the first and second cube, respectively, after the collision.
Since the collision is elastic, kinetic energy is conserved as well. Therefore, you can also use the conservation of kinetic energy equation in addition to the conservation of linear momentum equation to solve for the final velocities.
(1/2) * M1 * V0^2 = (1/2) * M1 * V1'^2 + (1/2) * M2 * V2'^2
Now you have two equations and two unknowns, V1' and V2'. To solve the system of equations, you can substitute V2' = 0.5 * V1' (since M2 is half the mass of M1), and solve for V1'.
Once you have found V1', you can substitute the values into the equation:
V0 = 2.43 m/s (as per your calculation)
Then, you can determine the distances each cube will land. Since both cubes are starting from rest, their vertical displacements will depend on the time it takes them to reach the ground level (90 cm off the floor).
Using the equation of motion:
h = (1/2) * g * t^2
where h is the vertical displacement (30 cm) and g is the acceleration due to gravity, you can solve for the time it takes for each cube to fall.
Once you have the time, you can calculate the horizontal displacement by multiplying the time by the horizontal velocity, V1' or V2' (since V2' = 0.5 * V1').
Now you have the necessary steps to solve the problem and find where each cube will land.