A snooker ball of mass 0.30kg slides on a frictional horizontal snooker board. Two snooker sticks strike the ball simultaneously.if the force F1 has a magnitude of 5N and points 20 South of East and the F2 has a magnitude of 8 N and points 60 north east. determine..
(I) the magnitude of the balls acceleration
(ii) the direction of the balls acceleration
TYPOS?
A snooker ball of mass 0.30kg slides on a frictional (FRICTIONLESS???) horizontal snooker board. Two snooker sticks strike the ball simultaneously.if the force F1 has a magnitude of 5N and points 20 South of East and the F2 has a magnitude of 8 N and points 60 north (OF ???) east. determine.
Force East = Fe = 8 cos 60 + 5 cos 20 = 4+ 4.70 = 8.70
ForceNorth= Fn = 8 sin 60 - 5 sin 20 = 6.93 - 1.71 = 5.22
magnitude = sqrt ( 8.70^2 + 5.22^2)
tangent of angle north of east = Fn/Fe = 5.22/8.70
To determine the magnitude and direction of the ball's acceleration, we need to break down the forces acting on the ball into their respective components.
First, let's break down the force F1 into its x and y components:
F1x = F1 * cosθ1
F1y = F1 * sinθ1
where θ1 is the angle of F1, which is 20 degrees south of east.
Similarly, let's break down the force F2 into its x and y components:
F2x = F2 * cosθ2
F2y = F2 * sinθ2
where θ2 is the angle of F2, which is 60 degrees north-east.
Next, let's compute the net force acting on the ball by summing up the x and y components:
Fnetx = F1x + F2x
Fnety = F1y + F2y
Now, we can calculate the net force magnitude:
|Fnet| = sqrt(Fnetx^2 + Fnety^2)
This magnitude represents the magnitude of the ball's acceleration (a = |Fnet| / m), where m is the mass of the ball.
Finally, to determine the direction of the ball's acceleration, we can use the following equation:
θa = arctan(Fnety / Fnetx)
Now, let's substitute the given values and solve for the required quantities:
θ1 = 20 degrees (south of east)
θ2 = 60 degrees (north-east)
F1 = 5 N
F2 = 8 N
m = 0.30 kg
Calculating the x and y components of F1:
F1x = 5 * cos(20) ≈ 4.64 N
F1y = 5 * sin(20) ≈ 1.71 N
Calculating the x and y components of F2:
F2x = 8 * cos(60) ≈ 4 N
F2y = 8 * sin(60) ≈ 6.93 N
Calculating the net force components:
Fnetx = F1x + F2x ≈ 8.64 N
Fnety = F1y + F2y ≈ 8.64 N
Calculating the net force magnitude:
|Fnet| = sqrt(Fnetx^2 + Fnety^2) ≈ 12.22 N
Calculating the ball's acceleration:
a = |Fnet| / m ≈ 12.22 / 0.30 ≈ 40.73 m/s^2
Calculating the direction of the ball's acceleration:
θa = arctan(Fnety / Fnetx) ≈ arctan(8.64 / 8.64) ≈ 45 degrees
Therefore,
(i) The magnitude of the ball's acceleration is approximately 40.73 m/s^2.
(ii) The direction of the ball's acceleration is approximately 45 degrees.