In the figure, point P is at rest when it is on the x-axis. The linear speed of point P when it reaches the y-axis is closest to
A) 0.18 m/s.
B) 0.24 m/s.
C) 0.35 m/s.
D) 0.49 m/s.
E) 0.71 m/s.

Question

In the figure, point P is at rest when it is on the x-axis. The linear speed of point P when it reaches the y-axis is closest to 
A) 0.18 m/s. 
B) 0.24 m/s. 
C) 0.35 m/s. 
D) 0.49 m/s. 
E) 0.71 m/s.

Quizplus · Accepted Answer

The key to solving this problem is conservation of energy. When point P is on the x-axis, it has only potential energy due to its height above the origin. As it moves towards the y-axis, the potential energy is converted into kinetic energy, and we can equate the two:

mgh = (1/2)mv^2

where m is the mass of point P, g is the acceleration due to gravity, h is the initial height of P above the origin (i.e. the distance from the x-axis), and v is the speed of P when it reaches the y-axis.

We know that the distance from the x-axis to the origin is 5 m, so the initial height of P is h = 5 sin(30°) = 2.5 m. We also know that the mass of P is given as 1 kg.

Plugging in these values and solving for v, we get:

(1/2)(1 kg)v^2 = (1 kg)(9.8 m/s^2)(2.5 m)

v^2 = 4.9 m^2/s^2

v ≈ 2.2 m/s

However, we have to remember that the question is asking for the linear speed, not the magnitude of the velocity vector. When P reaches the y-axis, its velocity vector is directed at a 45° angle to the y-axis (since it was moving at an angle of 30° to the x-axis). Therefore, the x-component of its velocity is v cos(45°) and the y-component is v sin(45°), and the linear speed is given by the Pythagorean theorem:

linear speed = sqrt[(v cos(45°))^2 + (v sin(45°))^2]

= sqrt[(v^2/2) + (v^2/2)]

= sqrt[v^2]

= v

Plugging in v = 2.2 m/s, we get a linear speed of approximately 0.35 m/s, which is closest to choice C.

In the Figure, Point P Is at Rest When It