Conservation of Momentum/Mechanical Energy Question (Fairly Challenging)

In summary, the small block of mass m is moving on a horizontal table surface at initial speed v0 and then moves onto a sloped big block of mass M without friction. The speed v of the small block after it leaves the slope can be found using the conservation of momentum equation, (M + m)v = -mv1 + Mv2, and conservation of kinetic energy equation, 1/2(M+m)v^2 = 1/2mv1^2 + 1/2Mv2^2, after eliminating v2. The height that it rises to is not relevant for this problem.
  • #1
Quotexon
11
0

Homework Statement


A small block of mass m is moving on a horizontal table surface at initial speed v0. It then moves smoothly onto a sloped big block of mass M. The big block can also move on the table surface. Assume that everything moves without friction
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Find the speed v of the small block after it leaves the slope.

Homework Equations


Conservation of Momentum: mv0 = (M+m)v → v=(mv0)/(M+m)
Conservation of Mechanical Energy: fairly obvious
The height that it rises to is h = (1/2g)(Mv0^2/M+m), I derived this from conservation of momentum and conservation of mechanical energy

The Attempt at a Solution


I set up the conservation of momentum: (M + m)v = -mv1 + Mv2 Is this correct?
Now I ask, how should I set up the conservation of mechanical energy equation? Should I solve for one variable in terms of the other? The answer should be in terms of v0.
The answer is v1 = ((M-m)/(M+m))v0
 
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  • #2
Quotexon said:
Conservation of Momentum: mv0 = (M+m)v → v=(mv0)/(M+m)
Don't assume that m & M have the same velocity.
Conservation of Mechanical Energy: fairly obvious
The height that it rises to is h = (1/2g)(Mv0^2/M+m), I derived this from conservation of momentum and conservation of mechanical energy
Don't worry about the height. That's an intermediate point that we don't care about.

The Attempt at a Solution


I set up the conservation of momentum: (M + m)v = -mv1 + Mv2 Is this correct?
Set initial momentum equal to final momentum. Let v1 and v2 be the final velocities after they separate.
Now I ask, how should I set up the conservation of mechanical energy equation?
Intial KE = final KE
Should I solve for one variable in terms of the other? The answer should be in terms of v0.
You'll have two equations. Eliminate v2 and solve for v1.
 
  • #3
How would a conservation of kinetic energy equation look like?

1/2(M+m)v^2 = 1/2mv1^2 + 1/2Mv2^2 ?

Conservation of momentum: (M+m)v = -mv1 + Mv2 ?

If I solve for v1 using these equations, it turns out very convoluted and does not resemble the answer. I feel as though it should be cleaner than this. Am I doing something incorrectly?
 
  • #4
Does this problem even meet the criteria to use conservation of kinetic energy? It doesn't seem to be elastic, does it?

Please help, this problem has been bugging me for awhile.
 
  • #5
Quotexon said:
How would a conservation of kinetic energy equation look like?

1/2(M+m)v^2 = 1/2mv1^2 + 1/2Mv2^2 ?
OK, except for the left hand side. Initially, only m is moving.

Conservation of momentum: (M+m)v = -mv1 + Mv2 ?
Same issue as above.

If I solve for v1 using these equations, it turns out very convoluted and does not resemble the answer. I feel as though it should be cleaner than this. Am I doing something incorrectly?
Clean up the equations as I suggest and try again. A bit of a pain, but you'll get the required answer with a bit of work.
 
  • #6
Wow, it worked! Thanks so much for your help. I greatly appreciate it.
 
  • #7
Excellent! :approve: (And you are most welcome.)
 

1. What is the principle of conservation of momentum?

The principle of conservation of momentum states that the total momentum of a closed system remains constant. This means that in the absence of external forces, the total momentum before a collision or interaction is equal to the total momentum after the collision or interaction.

2. How is the conservation of momentum related to collisions?

In collisions, the total momentum before the collision is equal to the total momentum after the collision. This is because in a closed system, the total momentum remains constant. Therefore, the momentum of each object involved in the collision will change, but the total momentum of the system will remain the same.

3. Is the conservation of momentum applicable to all types of collisions?

Yes, the conservation of momentum is applicable to all types of collisions, including elastic and inelastic collisions. In an elastic collision, both kinetic energy and momentum are conserved, while in an inelastic collision, only momentum is conserved.

4. What is the equation for calculating momentum?

The equation for momentum is: momentum = mass x velocity. This means that the momentum of an object is directly proportional to its mass and velocity.

5. How does the conservation of mechanical energy relate to the conservation of momentum?

The conservation of mechanical energy also follows the principle of conservation of momentum. In a closed system, mechanical energy (the sum of kinetic and potential energies) is conserved, meaning that the total mechanical energy before a collision is equal to the total mechanical energy after the collision. This is because, in the absence of external forces, the total momentum and total mechanical energy of a closed system remain constant.

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