Kinematics on a frictionless incline plane

[Eyris]

The situation:
The mass is pushed up an incline with an angle of 1 degree with an initial velocity of 1 m/s, and it comes back down to its original position.

The questions to answer:
What is the total distance the object travels on the frictionless inclined plane?
How long will the object take to return to its starting position?

Relevant equations:
V = Vi + at
V^2 = V^2i + 2a(X-Xi)

Knowns:
Angle = 1 degree
Initial Velocity Up Plane = 1 m/s
Final Velocity at top of plane = 0 m/s
a = g*sin(angle)

My attempt:
My positive X axis is pointing up the incline, and my positive Y is perpendicular to the mass in the up direction.

What is the total distance?
d = (X-Xi)

The kinematic equation I will use to find the distance from bottom to top of incline:
V^2 = V^2i + 2 * a * d

a = -9.81 * sin(1degree) = -.171 m/s^2
V = 0 m/s (at the top of incline, velocity is 0)
Vi = 1 m/s

Plug in a, v, and vi into v^2 equation to solve for d:
0 = 1 + 2(-.171)*d
d = 2.92m (to the top)
2d = total distance

What is the total time?
total time = time to top + time to bottom

To the top
V = Vi + at
a = -.171 m/s^2
t = (V-Vi)/a = -1/-.171 = 5.85s (to the top)

To the bottom
Vi = 0 (at top of plane)
d = 2.92m
a = +.171 m/s^2
V^2 = 0 + 2(.171)(2.92) -- (Am I proceeding correctly?)
V = 1 m/s (would this actually equal -1; if so, is my previous V^2 equation incorrect?) V WHEN IT COMES BACK TO ITS ORIGINAL POSITION
V = Vi + at
1 = 0 + (.171)(t)
t = 1/.171 = 5.85s (time to go down)

total time = 5.85 * 2 = 11.7s

 

[gneill]

Hi Eyris,

Welcome to Physics Forums!

Your calculations and results look good.

Note that you might also have used energy conservation (KE versus gravitational PE) to find the height the mass achieved, and hence the distance along the plane with a bit of trig (you have the angle and the opposite side). You could also have employed the basic trajectory equation,

$s = v_ot + \frac{1}{2}a t^2$

to find the total time, since the mass returns to its starting point, so just solve for t when s = 0.

 

[Eyris]

Hi Eyris,

Welcome to Physics Forums!

Your calculations and results look good.

Note that you might also have used energy conservation (KE versus gravitational PE) to find the height the mass achieved, and hence the distance along the plane with a bit of trig (you have the angle and the opposite side). You could also have employed the basic trajectory equation,

$s = v_ot + \frac{1}{2}a t^2$

to find the total time, since the mass returns to its starting point, so just solve for t when s = 0.

I appreciate the confirmation. I have one last question. Would the velocity, once the object has returned from moving up the inclined plane, be negative or positive? It came back down in the direction of the negative x-axis as define by choice of x and y.

Also, the basic trajectory equation would have worked too! In that case, I would have used 2d (2 times the distance from the initial position to the top) to represent the total distance overall, I believe. Correct me if I am wrong. My concern, however, was that the time coming back down would be different from the time going up, so I divided it into two pieces. If that is not the case, might you explain the reasoning behind that being false, if you don't mind?

 

[gneill]

I appreciate the confirmation. I have one last question. Would the velocity, once the object has returned from moving up the inclined plane, be negative or positive? It came back down in the direction of the negative x-axis as define by choice of x and y.

The velocity, upon return, would be directed in the opposite direction to the initial velocity. If your choice of coordinates made the initial velocity positive, then yes, upon return the velocity would be negative.

Also, the basic trajectory equation would have worked too! In that case, I would have used 2d (2 times the distance from the initial position to the top) to represent the total distance overall, I believe. Correct me if I am wrong. My concern, however, was that the time coming back down would be different from the time going up, so I divided it into two pieces. If that is not the case, might you explain the reasoning behind that being false, if you don't mind?

Without friction the trajectory will be symmetric (in terms of time) going up and coming down. So the time from launch to maximum height (or distance along the plane) will be equal to the time to return from that turnaround point.