Magnitude of a ball's Acceleration

[PhysicsHelp152]

Homework Statement

A 0.17-kg cue ball rests on the pool table. It's struck by a cue stick applying force F1→ = 13Ni^+21Nj^
Determine the magnitude of the ball's acceleration.

Homework Equations

a_x = F_{net, x} / m
a_y = F_{net, y} / m
a = F_net / m

The Attempt at a Solution

Components of Acceleration:
a_x = 13Ni^ / 0.17kg = 76.47m/s²

a_y = 21Nj^ / 0.17kg = 123.53m/s²

Magnitude:
F_net = √(13Ni^)² * (21Nj^)² = 24.7N

angle θ = tan^-1 * (21N / 13N) = 58.24° above the + x-axis (fixed. see replies below)

Acceleration's Magnitude:
a = F_net / m = 24.7N / 0.17kg = 145.29m/s² or 145m/s² rounded

I feel like this should be the answer, or that I'm very close to the answer. Am I missing a step? Did I do something wrong? Am I way off?

 

[gneill]

Hi PhysicsHelp152, Welcome to Physics Forums.

Your work and results look fine except for the angle. Take another look at which component is which for the arctan function to find the angle w.r.t. the x-axis.

 

[PhysicsHelp152]

Is this the correct angle?
θ = tan-1 * (21N / 13N) = 58.24° (fixed. see replies below)

Also, the magnitude of the ball's acceleration is indeed 145m/s²?
I ask because apparently it isn't correct? I answer my questions online and I get 5 tries to get it right. I only have one try left and I already tried 145m/s².

 

[gneill]

Is this the correct angle?
θ = tan-1 * (21Ni^ / 13Nj^) = 58.24°

Yes, that looks better. But don't include the axis unit vectors in the formula: you're taking a ratio of component values, not vector quantities. Besides, the x-component is generally designated by the $\hat{i}$ unit vector, and the y-component by the $\hat{j}$ unit vector, and you seem to have swapped the components of the force vector by using "21Ni^" and "13Nj^".

Also, the magnitude of the ball's acceleration is indeed 145m/s²?
I ask because apparently it isn't correct? I answer my questions online and I get 5 tries to get it right. I only have one try left and I already tried 145m/s².

Your calculation looks fine. You may be getting hit by significant figures. How many sig figs in the given information?

 

[PhysicsHelp152]

Your calculation looks fine. You may be getting hit by significant figures. How many sig figs in the given information?

There are 2 sig figs in the given information... I think. So my answer should be two sig figs?
But what would the answer be then?