So momentum in the y direction should cancel out?
4.33 M sin30 = MV sin θ
And the momentum in the x direction will equal the original momentum so
5M = 4.33 M cos30 + MV cosθ
I'm guessing this question can be solved using the law of conservation of momentum
Vi = 5 m/s
(5 m/s) M = (4.33 m/s) cos30 M + V sinθ M
I don't know what to do after this... I'm also not sure if I use the sin and cos correctly.
First of all, the pulling force is
300N cos(30) = 260 N
At this point, I try to find the friction force
Fn = mg = 20kg * 9.81 m/s^2 = 196.2 N
Then,
Ff = μ * Fn = 0.5 * 196.2 N = 98.1 N
So after canceling the horizontal forces,
260N - 98.1N = 161.9N
And the acceleration will be 161.9N / 20kg =...