Recent content by fantisism

Alright. I can solve the second subquestion myself. Thank you again!

To find the angle, would it be arctan(v2fy/v2fx)?

Oh alright then. I thought v in the n-mg=m(v2/r) was the change in velocity, although now that I think about it, it's unnecessary. Thank you so much!

Oh sorry. I just wanted to make sure those were right because I tend to make small simple math errors. Thank you!

And then once I find that out, I would subtract vi from vf and use that value for v in n-mg=m(v2/r)?

vf=sqrt(2gh+(vi)2)?

And then once I figure out what those two velocities are, I would square them individually and take the square root of the sums? So, sqrt((v2fx)2+(v2fy)2? And that would be the answer for the first question?

So if I had to solve vf, it would be sqrt(2gh)+3 m/s?

And then with that information, solve for v2fx and v2fy?

So in order to find the velocity when the ball reaches the bottom of the loop, I would use 0.5*m*(vi)2+0.5*m*(vf)2=mgh and solve for vf?

Is v1fx=(4 m/s)(cos(25°)) and v1fy=(4 m/s)(sin(25°))?

Would I use v=sqrt(2gh) in this example?

Homework Statement You slide a 2 kg block on a flat icy surface towards a static block of mass M. Before the collision, the 2 kg mass is moving at 3 m/s. The total kinetic energy is not changed by the collision. After the collision, the 2 kg mass is moving at −1 m/s i.e. opposite to its...

Homework Statement Two ice skaters crash into each other. Before they collide, one of them (50 kg) is skating in a straight line at 5 m/s, the other (40 kg) is skating at 4 m/s in a straight line at 90 degrees to the 50 kg skater’s direction. After the collision, the 50 kg skater is moving at 4...

Homework Statement Two lumps of clay, each with the same mass 3.5 kg, are sliding on ice, with no friction. One is moving at 4.8 m/s to the right, the other is moving at 2.3 m/s to the left. They collide and stick to one another. Calculate the velocity of the lump of clay after the collision...