Velocity after collision and Kinetic Energy.

AI Thread Summary
In this discussion, the principles of momentum conservation and kinetic energy are explored through a collision scenario involving two trolleys. The total momentum before the collision is calculated to be 3 kg·m/s, and the velocity of the second trolley after the collision is determined to be 2 m/s. It is established that the collision is inelastic since kinetic energy is lost, with the initial kinetic energy being greater than the final kinetic energy. The concept of "rebound" is clarified, emphasizing that it refers to a change in direction of velocity, which is crucial in momentum calculations. The importance of treating momentum as a vector quantity is highlighted, as direction affects the overall momentum after the collision.
Meezus
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Homework Statement


A trolley of mass 0.5 kg rolls along a level, frictionless ramp at a velocity of 6.0 m s-1. It collides with a second trolley of mass 1.0 kg which is initially at rest. The first trolley rebounds with a speed of 2.0 m s-1 .
a) Stating clearly any principles involved, find the velocity of the second trolley after the collision.
b) Compare the kinetic energy before and after the collision. Is the collision elastic?

Homework Equations


p = m x v
E=1/2*m*v^2

The Attempt at a Solution


Momentum before collusion for carriage A: 0.5 x 6 = 3
Momentum before collusion for carriage B: 1 x 0 = 0
Total Momentum before : 3 + 0 = 3.
Total Mass = 1 + 0.5 = 1.5.
p = m x v
v = p / m
3 / 1.5 = 2.
The principles involved are that as long as no external forces are acting on the object the total momentum stays the same in a collisions.
Is this correct for the first one?

part b:[/B]

0.5 x 1.5 x 62 = 27
0.5 x 1.5 x 22 = 3

The collision is inelastic as kinetic energy is lost?
 
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What is meant by "rebounds"?
 
NascentOxygen said:
What is meant by "rebounds"?

Like when you throw a ball at a wall and it comes back. It "rebounds" off wall.
 
So the two bodies do not stick together.

Then try to complete this table, too:

Momentum after collision for carriage A:
Momentum after collision for carriage B:
Total Momentum after collision:
 
NascentOxygen said:
So the two bodies do not stick together.

Then try to complete this table, too:

I thought because momentum is conserved the total momentum is the same after?
 
Meezus said:
I thought because momentum is conserved the total momentum is the same after?
It should be. Can you fill in the blanks?
 
NascentOxygen said:
It should be. Can you fill in the blanks?

Momentum after collision for carriage A: 0.5 x 2 = 1
Momentum after collision for carriage B:1 x ? = ?
Total Momentum after collision: 3

If the total momentum stays the same the velocity for carriage B must be 2.
 
If a body rebounds, its velocity changes direction.
 
NascentOxygen said:
If a body rebounds, its velocity changes direction.

Ok. Is what I have done correct? I don't see where the rebound part comes into play?
 
  • #10
Meezus said:
Ok. Is what I have done correct? I don't see where the rebound part comes into play?
Velocity and momentum are vectors. Direction matters. If a ball hits a wall square on at velocity v and rebounds with the same speed, what is its new velocity?
 
  • #11
total momentum is conserved.
total momentum before collision = total momentum after collision
(if there is no external unbalanced force is acting there! )
 
  • #12
Meezus said:
Momentum after collision for carriage A: 0.5 x 2 = 1
The correct answer is 0.5 x -2 because rebound velocity is in the opposite direction. Momentum is a vector quantity.
 
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