Recent content by Daniel Guh

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.

Ohh, do I have to include the force from the 300N? Since there is a downward force, the normal force will be greater?

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 =...