A 5-g marble, released form rest, was allowed to roll on a ramp whose lower end is 25 cm from the floor. When it left the ramp, it behaved like a body in projectile motion. What was the horizontal velocity of this marble when it left the ramp? With what momentum did it leave the ramp?

You have to know something about the ramp to answer these questions.

M= 1525g= 1.525 kg

W= 14.95 N

Force required to start the block into motion:
F= 0.685(9.8)= 6.71N

Force to maintain motion:
F= 0.512(9.8)= 5.02N

Compute µk and µs:
µk = 5.02/14.95 = 0.34 µs = 6.72/14.95 = 0.45

To find the horizontal velocity of the marble when it left the ramp, we can use the principles of projectile motion.

First, we need to find the time it takes for the marble to reach the floor after leaving the ramp. We can use the equation for vertical displacement:

h = (1/2) * g * t^2

Given that the height is h = 25 cm = 0.25 m and acceleration due to gravity is g ≈ 9.8 m/s², we can rearrange the formula to solve for time:

t = √(2h / g) = √(2 * 0.25 / 9.8) ≈ 0.223 s

Now, considering the horizontal motion, we can use the formula for constant velocity:

v = d / t

Where v is the velocity, d is the horizontal distance traveled, and t is the time. In this case, we need to find the horizontal velocity, so the formula becomes:

v = d / t

The horizontal distance traveled by the marble when it left the ramp is the same as the distance from the ramp to the floor. Let's assume this is denoted by "d".

Therefore, the horizontal velocity (v) of the marble when it left the ramp can be calculated as:

v = d / t
v = d / 0.223

To find the value of "d," we need more information. Please provide the length of the ramp or any other relevant measurements so I can assist you further.