A small cube of mass m1= 1.0 kg slides down a circular and frictionless track of radius R= 0.4 m cut into a large block of mass m2= 4.0 kg as shown in the figure below. The large block rests on a horizontal and frictionless table. The cube and the block are initially at rest, and the cube m1 starts from the top of the path. Find the speed of the cube v1 as it leaves the block. Take g= 10.0 m/s2. Enter your answer in m/s.

Question

A small cube of mass m1= 1.0 kg slides down a circular and frictionless track of radius R= 0.4 m cut into a large block of mass m2= 4.0 kg as shown in the figure below. The large block rests on a horizontal and frictionless table. The cube and the block are initially at rest, and the cube m1 starts from the top of the path. Find the speed of the cube v1 as it leaves the block. Take g= 10.0 m/s2. Enter your answer in m/s.

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Answer 1

To find the speed of the cube (v1) as it leaves the block, we can use the principle of conservation of mechanical energy.

When the cube starts from the top of the track, it has potential energy which converts into kinetic energy as it slides down. At the bottom of the track, this kinetic energy is transferred to the block, causing it to move.

We can set up the equation for energy conservation as follows:

Initial potential energy of the cube = Final kinetic energy of the cube + Final kinetic energy of the block

For the cube:
Initial potential energy of the cube = m1 * g * h
Final kinetic energy of the cube = 0.5 * m1 * v1^2

For the block:
Final kinetic energy of the block = 0.5 * m2 * V^2

As there is no friction, energy is conserved, so the sum of the initial potential energy of the cube and the final kinetic energy of the block should be equal to each other.

m1 * g * h = 0.5 * m1 * v1^2 + 0.5 * m2 * V^2

Solving for v1, we get:

v1 = sqrt((2 * (m1 * g * h - 0.5 * m2 * V^2)) / m1)

Given:
m1 = 1.0 kg (mass of the cube)
m2 = 4.0 kg (mass of the block)
R = 0.4 m (radius of the track)
g = 10.0 m/s^2 (acceleration due to gravity)

To find h (height of the track), we need to use the radius R:
h = R - R = 0 m

Now we can substitute the values into the equation and find v1:

v1 = sqrt((2 * (m1 * g * h - 0.5 * m2 * V^2)) / m1)
= sqrt((2 * (1.0 kg * 10.0 m/s^2 * 0 - 0.5 * 4.0 kg * V^2)) / 1.0 kg)
= sqrt((2 * (- 8.0 kg * V^2)) / 1.0 kg)
= sqrt(-16.0 V^2) m/s

Since the result is the square root of a negative value, it means that the cube does not leave the block. It means that the height of the track is not enough for the cube to transfer its kinetic energy to the block.

Therefore, the cube does not leave the block and the speed v1 is 0 m/s.