Momentum and energy questions

[Farside]

Should be fairly simple, but I'm not so great with physics. Here's a few I'm having trouble with...



1. Under what circumstances (if any) is no work done on a moving object even though a net force acts upon it?

2. What happens to the momentum of a car when it stops?

3. When the kinetic energy of an object is doubled, what happens to its momentum?

4. A railway car is at rest on a frictionless track. A man at one end of the car walks to the other end. (a) Does the car move while he is walking? ( If so, in which direction? © What happens when the man comes to a stop?

 

[christinono]

Should be fairly simple, but I'm not so great with physics. Here's a few I'm having trouble with...



1. Under what circumstances (if any) is no work done on a moving object even though a net force acts upon it?

2. What happens to the momentum of a car when it stops?

3. When the kinetic energy of an object is doubled, what happens to its momentum?

4. A railway car is at rest on a frictionless track. A man at one end of the car walks to the other end. (a) Does the car move while he is walking? ( If so, in which direction? © What happens when the man comes to a stop?

Ok, think of the formula for work. W=Fdcos(theta), right? For what value of theta does cos(theta) equal zero.

For the second question, again, think of the formula for momentum.
Momentum = mv. No velocity, no momentum.

For the third one, think of the formula for kinetic energy. What is the relationship between Ek and velocity? When the kinetic energy is doubled, what happens to the velocity? Once you know this relationship, you the the relationship between kinetic energy and momentum, since momentum is directly related to velocity.
For the fourth question, use the conservation of momentum concept.

 

[Farside]

Okay, I still don't quite get the first one. The material wasn't presented in that way. Are there no circumstances?

 

[illusion88]

1:
its simple.
work done =F.s cos(theta)
so when theta=90 then cos(theta)=0.i.e.when the force is applied perpendicular to the direction of motion of a body,no work will be done.

2.momentum =mv
therefore as soon as the car stops(v=0) the momentum changes to 0.

3.Momentum=root(2k.e.*m)
so when k.e. is doubled momentum is increased by a factor of root(2).

4.This question is a bit tricky.one thing has to be clarified.if there is no friction between the man and the railroad car then he cannot move on it.

Sriram