Inelastic collisions - conservation of energy

[mintsnapple]

Homework Statement

Homework Equations

K = E + U
E = 1/2mv^2
U = mgh

The Attempt at a Solution

So I think that for a, half the final velocity of m is equal to the initial velocity of 2m. So that's v = sqrt(2gr)/2? Momentum is not conserved because normal force and gravity are acting on block m correct? Then what do I do from there, and on b and c?

 

[SammyS]

Homework Statement

Homework Equations

K = E + U
E = 1/2mv^2
U = mgh

The Attempt at a Solution

So I think that for a, half the final velocity of m is equal to the initial velocity of 2m. So that's v = sqrt(2gr)/2? Momentum is not conserved because normal force and gravity are acting on block m correct? Then what do I do from there, and on b and c?

This is a multi-stage problem.

Attack it systematically.

First, the block with mass, m, slides down the ramp. What is its velocity immediately before it collides with the block of mass 2m ?

Then there is the collision.

What velocity does each block have immediately after the collision?

The rest of this depends upon those velocities.

 

[mintsnapple]

So then,
mgr = 1/2mv1 + 1/2(2m)v2 right?
So, v2 = sqrt(gr-1/2(v1)^2)

From the kinematics formula, the time it takes for the block to fall is
R = 1/2gt^2
So t = sqrt(2R/g)

But my book says the answer is 4/3*sqrt(Rr), but v2*t is clearly not the same...what am I doing wrong?

 

[SammyS]

So then,
mgr = 1/2mv1 + 1/2(2m)v2 right?
So, v2 = sqrt(gr-1/2(v1)^2)

From the kinematics formula, the time it takes for the block to fall is
R = 1/2gt^2
So t = sqrt(2R/g)

But my book says the answer is 4/3*sqrt(Rr), but v2*t is clearly not the same...what am I doing wrong?

Which answer?


It appears that you have not taken care to go though each step that's involved here.

You do have the speed of the small mass immediately prior to the collision correct.


Then there is the collision.

How do you propose to analyze that? It's an elastic collision. What velocity does each mass have immediately after the collision (including direction).

Hint: Yes, you need to use conservation of momentum.​

After that, you might begin to concern yourself with the distances asked for.


After that you

 

[HalfLostAndSearching]

Inelastic collisions involve a transfer of kinetic energy between objects, resulting in a decrease in the total kinetic energy of the system. However, the conservation of energy principle still applies, as the total energy (kinetic plus potential) remains constant throughout the collision. This can be seen through the equation K = E + U, where K is the total kinetic energy, E is the total energy, and U is the total potential energy. In the case of an inelastic collision, the decrease in kinetic energy is balanced by an increase in potential energy.

In terms of the homework problem, it is important to recognize that the final velocity of m is not simply half of the initial velocity of 2m. The final velocity will depend on the specific circumstances of the collision, such as the masses and velocities of the objects involved. Additionally, you are correct in noting that momentum is not conserved in an inelastic collision, as external forces such as normal force and gravity are acting on the objects.

To solve the problem, you can use the conservation of energy principle to set up equations for the initial and final energies, and then solve for the final velocity of m. For part b and c, you can use the same approach, but with different initial conditions and equations specific to the given scenarios. It is important to carefully consider all the forces and energies involved in each collision to accurately solve the problem.