What Happens to the Velocities in a Frictionless Curved Surface System?

In summary, the conversation discusses a problem involving a small mass sliding on a larger mass and breaking off at a certain height. The first part of the problem is solved using the principle of conservation of linear momentum, and the second part is solved by considering the entire system and applying the law of conservation of energy. The conversation also briefly touches on the concept of frames of reference and their impact on kinetic energy calculations.
  • #1
Nikhil Kumar
6
0
A small body of mass m placed over a larger mass M whose surface is horizontal near the smaller mass and gradually curves to become vertical. The smaller mass is pushed on the longer one at a speed v and the system is left to itself. Assume all surfaces to be frictionless. (Please refer the attached diagram).

Diagram.JPG


a) Find the speed of the larger block when the smaller block is sliding on the vertical part.

b) Find the speed of the smaller mass when it breaks off the larger mass at height h.

2. Homework Equations

Principle of conservation of linear momentum
Law of conservation of energy

3. The Attempt at a Solution

I have correctly solved the first part of the question. Here it goes:
Let velocity of larger block (M) when smaller block (m) reaches the vertical part be V, then by conservation of linear momentum ( I've taken the smaller mass + larger mass as system):

Initial linear momentum of system = final linear mom. of system
» mv + 0 = (M + m)V
» V = mv / (M+m) (This is correct)

However, i am stuck at the second part. I tried it attempting like this:

Taking the smaller mass as the system, initial energy of the system= ½ mv2
Let v1 be the vel. of the smaller mass ( with respect to ground) when it breaks off from the vertical mass, then total energy at this instant = ½ mv12 + mgh (assuming PE = 0 at ground level)

As no external force acts on the system, the mech. energy is conserved, hence:
Intial Mechanical energy = Final Mechanical energy
» ½ mv2 = ½ mv12 + mgh
» v1 = [v2 - 2gh]1/2

But this is wrong! What is the fallacy in my logic?

Another allied question: Suppose a frame of reference is moving with a const. velocity v wrt ground. A body of mass m moves with a const. vel. V wrt the frame of reference. then what is its kinetic energy? Which vel. should we take for calculating KE? Like PE, is KE also relative?

Thanks.
 
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  • #2
Can you take the smaller mass "as the system"? Try looking at the whole system, including the larger mass before and after...
 
  • #3
Thanks sir! This has solved my problem. I solved it like this:
Let v1 be the velocity of the smaller mass wrt the ground when it just flies away from the larger mass. Taking 'the larger mass + smaller mass as the system' and applying law of cons. of energy, we have:

Intial energy = Final energy
½ mv2 = ½ MV2 + ½ mv12 + mgh

Hence, v1 = {[(m2 + Mm + m2)/(M+m)2]-2gh}1/2

Thanks sir.
 
  • #4
Nikhil Kumar said:
Another allied question: Suppose a frame of reference is moving with a const. velocity v wrt ground. A body of mass m moves with a const. vel. V wrt the frame of reference. then what is its kinetic energy? Which vel. should we take for calculating KE? Like PE, is KE also relative?

Thanks.

What is the answer to this question?
 
  • #5
Nikhil - thanks you for introducing an interesting problem and doing most of the hard work. And don't call me sir! It makes me feel old :wink: Just Mal will do.

On "the frame of reference" thing. If K is directly proportional to v squared, and v is (obviously) relative then...
 
  • #6
Thats a question from HC Verma. Hey Nikhil, ar you Nikhil Kumar Singh from CMS Aliganj?
 
  • #7
Thanks Mal! You made my day.

And Ritwik, I'm not that Nikhil. I live in Chattisgarh.
 

FAQ: What Happens to the Velocities in a Frictionless Curved Surface System?

What is conservation of linear momentum?

The law of conservation of linear momentum states that the total momentum of a closed system remains constant over time, as long as there are no external forces acting on the system. This means that the total momentum before a collision or interaction is equal to the total momentum after the collision or interaction.

How is linear momentum calculated?

Linear momentum (p) is calculated by multiplying the mass (m) of an object by its velocity (v). This can be represented by the equation p = m * v. The unit of momentum is kilogram meters per second (kg*m/s).

What is an example of conservation of linear momentum in action?

An example of conservation of linear momentum is a game of pool. When the cue ball hits another ball, the total momentum before the collision (equal to the momentum of the cue ball) is equal to the total momentum after the collision (equal to the combined momentum of all the balls). This is true even if the balls change direction or stop moving after the collision.

Can linear momentum be negative?

Yes, linear momentum can be negative. This occurs when an object is moving in the opposite direction of its chosen reference frame. In this case, the momentum is still calculated by multiplying mass and velocity, but the velocity is given a negative value to indicate its direction.

How does conservation of linear momentum relate to Newton's third law?

Newton's third law of motion states that for every action, there is an equal and opposite reaction. This means that when two objects interact, they exert equal and opposite forces on each other. Conservation of linear momentum is a direct result of this law, as the total momentum of the system is always conserved due to these equal and opposite forces.

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