Question

An inclined plane P, 1.6 m long, makes an angle 30° with the horizontal. A point mass M is shot from the bottom of P at a point A with a speed Vo = 5 m/s up the plane of greatest slope. Neglect air resistance.

1. How long does it take M to reach B, the top of P? Then find its velocity.

2. M leaves the plane at B without meeting any obstacle to hit the horizontal plane containing A at a point C.

3. Determine the position of C. Give all necessary explanations​

Answer 1

The given length of the inclined plane of 1.6 m. inclined at 30° to the horizontal, and point mass velocity of 5 m/s gives;

1. 0.199 seconds
2. The distance traveled horizontally from the top is 0.086 m
3. 1.472 meters horizontally from A

How can the time and distance of the mass be found?

Length of the inclined plane = 1.6 m

Angle of inclination of the plane = 30°

Speed of the point mass = 5 m/s

1. Height of the top of P above ground, h is found as follows;

h = 1.6 m × sin(30°) = 0.8 m

The equation of motion that can be used is; h = u•t - 0.5•g•t^2

Which gives;

0.8 = 5•t - 0.5×9.81×t^2

0 = 5•t - 4.905×t^2 - 0.8

t = 0.199, or t = 0.821

• The time it takes the point mass to reach the top is approximately 0.199 seconds

2. At the top, we have;

Horizontal velocity, vx = 0.5 × cos(30°)

Which gives;

Horizontal velocity, vx = 0.433 m/s

The distance traveled, x is found as follows;

= 0.433 m/s × 0.199 s = 0.086 m

• The distance traveled by the point mass x = 0.086 m

3. The position of point C is therefore;

C(0.086 + 1.6 × cos(30°), 0)

C(1.472, 0)

Therefore;

• The position of point C is 1.472 meters from point A

Learn more about the equations of motion here:

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