Recent content by hyunxu

Yes , you're right!

We can consider the ball to be in rest initially so the initial velocity will be zero.

For finding momentum, you can use the formulaP=mv For finding final velocity , you can use , the kinematic equation v^2 = u^2+ 2as and to find that a , use a = f ÷m

Could you explain me? What is this radial component?(sorry because I'm absolute beginner but I want to learn it anyway)

But it is conserved unless an external force acting on it?In this case a sand bag is placed suddenly.Then there must be a force exerted by sandbag on the cart.

Oh Thank you for that! Thank you for guiding me in a correct way! I hope to learn a lot from you!

Thank you I got it!

Thank you for correcting my mistake.I have learned much from as a normal higher secondary school student!

What do we call force? In simple words force is a pull or push on object. we are standing on the surface of the earth. Because of a force exerted by it. So that force pulls us down. Hence we stand on the surface of the earth.Obviously this force is nothing but gravity. so I conclude that...

My question is simple. What's the role of static friction here?

Please explain this. How could you say it is static? So far I can imagine that the car is moving , it isn't at rest.Hence there's kinetic friction. From which frame of reference it is static?

why does static friction involve in this?since the object is moving.Why it doesn't have kinetic friction?

Let me explain it through simple equations N pull = mg - F cosθ N push = mg+F cosθ so these equations show that pulling is easier than pushing. But in practical life , I feel pushing is easier.So that's my question.

Could you please explain the term 'co-efficient of static friction'? why do the banking of roads or tracks depend of co-efficient of static friction?and not on the co-efficient of kinetic friction?

I guess pulling is much easier because total mass and energy required is less while comparing to pushing.