a)When the metal ball leaves the tray, the tray will be 10.78 cm above point A.
b) The total time between the system being released at point A and the ball leaving the tray is 0.086 seconds.
c) The speed of the ball as it exits the tray is 1.175 m/s.
Given,
Mass of tray , m_t=1.90 kg
spring of force constant, k = 200N/m
mass of the ball, m_b = 300 g
Height below the equilibrium point A = 16.4 cm
a)
The tray's height above point A when the metal ball leaves it- The total mass of the tray and ball is 2.2 kg (1.90kg+0.3 kg). The spring force is equal to the product of the spring constant and displacement.Thus,
f=kx
but we know that, f=mg
equating both the equations
[tex]mg=kx\\\\2.2*9.8=200*x\\\\21.56=200x\\\\x=\frac{21.56}{200}\\\\x=0.1078 m=10.78 cm[/tex]
As a result, the tray is pushed down 17.4 cm below its equilibrium point (point A) and released from rest. Thus, the tray's position above point A is h=16.4+10.78=27.18 cm.
b)
The total amount of time that passes between releasing the system at point A and the ball leaving the tray.
The formula for the simple harmonic function of a spring is as follows:
[tex]y=Acos(\sqrt{\frac{k}{m}}-\phi)[/tex]
Here,
y= displacement = 10.78 cm
A=amplitude i.e 16.4 cm
k= spring constant = 200 N/m
m= total mass = 2.2 kg
and [tex]\phi = 0 \degree[/tex]
[tex]10.78=16.4cos(\sqrt{\frac{200}{2.2}}t-0)\\\\\frac{10.78}{16.4}=cos(\sqrt{90.909}t)\\\\0.657=cos(9.534t)\\\\cos^{-1}(0.657)=9.5434t\\\\0.829=9.543 t\\t=\frac{0.829}{9.543}\\\\t=0.086 s[/tex]
c)
The speed at which the ball moves as it exits the tray
The total mechanical energy for simple harmonic motion of a spring can be given as,Here, take the distance measurements in meters
[tex]\frac{1}{2}mv_y^2+\frac{1}{2}ky^2=\frac{1}{2}kA^2\\\\v_y^2=k\frac{A^2-y^2}{m}\\\\v_y^2=200*\frac{0.164^2-0.1078^2}{2.2}\\\\v_y^2=90.909(0.0268-0.0116)\\\\v_y^2=1.38\\\\v_y=1.175 m/s[/tex]
Thus, the speed of ball is 1.175 m/s
To learn more about simple harmonic motion refer here
https://brainly.com/question/28208332
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