Inelastic collision: block moving down a frictionless ramp

In summary: So yes, the summary is correct.In summary, two suitcases, one with a mass of 11.8 kg and the other with a mass of 23.6 kg, are on a 6.36 m high frictionless ramp. The top suitcase is released and slides down the ramp, colliding with the second suitcase at rest. The collision is inelastic and the combined speed of the two suitcases can be found using the equation m1v1+m2v2= (m1+m2)v. The solution involves finding the velocities of each suitcase using the equations KE = (1/2)*mv2 and PE = mgh, and then using the inelastic collision formula to solve for the combined velocity
  • #1
Ly444999
19
0

Homework Statement


In the figure provided, two suitcases are on a 6.36 m high ramp to passengers waiting in a baggage terminal. The top suitcase is released from rest, and it slides down the ramp and hits the second suitcase. If the suitcase at the top has a mass of 11.8 kg and the other suitcase has a mass of 23.6 kg, what is their combined speed if an inelastic collision is achieved? Assume the ramp to be frictionless.

Homework Equations


KE = (1/2)*mv2
PE = mgh
m1v1+m2v2 = (m1+m2)v

The Attempt at a Solution


I just had a question about whether this is the approach to solve this question or not.
So since the ramp is frictionless, total mechanical energy is conserved.
You use mgh = (1/2)*mv2 , isolate v and solve for v for the first suitcase.
Then with the v you plug it into the inelastic collision formula to solve for the new combined velocity of the two suitcases.

m1v1+m2v2= (m1+m2)v
Isolate for v and v2 is 0?
 
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  • #2
I think it looks right to me. All the potential energy of the first suitcase gets converted to kinetic energy. That suitcase collides with the 2nd suitcase which is at rest. Since the collision is inelastic, there is conservation of momentum, but not conservation of energy.
 
  • #3
Inelastic collision so momentum transfer during collision is given by the equation Δp = μΔv where μ = reduced mass 11.8 x 23.6 /(11.8 + 23.6) and Δv = √(2gh).

Using g = 9.8 m/s2 we obtain Δp = 87.83 Ns and dividing by mass of suitcase 2, velocity is 3.72 m/s.

Now check on suitcase 1:

11.8 x √(2gh) - 87.83 = 43.92 and dividing by mass of suitcase 1 , velocity is also 3.72 m/s
 
  • #4
neilparker62 said:
Inelastic collision so momentum transfer during collision is given by the equation Δp = μΔv where μ = reduced mass 11.8 x 23.6 /(11.8 + 23.6) and Δv = √(2gh).

Using g = 9.8 m/s2 we obtain Δp = 87.83 Ns and dividing by mass of suitcase 2, velocity is 3.72 m/s.

Now check on suitcase 1:

11.8 x √(2gh) - 87.83 = 43.92 and dividing by mass of suitcase 1 , velocity is also 3.72 m/s
This is a homework forum. Please do not post complete solutions (unless it has already been solved and you are just showing a better way). Just point out errors in the posted attempt and offer hints on how to proceed.
In the present case, a simple "yes" would have been appropriate.
 
  • #5
Apologies - I was showing another method but that wasn't what was asked by the OP.
 

FAQ: Inelastic collision: block moving down a frictionless ramp

1. What is an inelastic collision?

An inelastic collision is a type of collision in which the kinetic energy of the objects involved is not conserved. This means that some of the kinetic energy is transformed into other forms of energy, such as heat or sound, and is not fully transferred to the objects after the collision.

2. How does a frictionless ramp affect an inelastic collision?

A frictionless ramp does not affect the inelastic collision itself, but it can affect the motion of the objects after the collision. Since there is no friction on the ramp, the objects will continue to move at a constant velocity after the collision, rather than slowing down due to friction.

3. What is the difference between an inelastic collision and an elastic collision?

In an elastic collision, the kinetic energy of the objects is conserved and there is no loss of energy. In an inelastic collision, some of the kinetic energy is lost and transformed into other forms of energy. Additionally, in an elastic collision, the objects bounce off of each other, while in an inelastic collision, they stick together or deform.

4. How is momentum conserved in an inelastic collision?

Momentum is always conserved in any type of collision, including inelastic collisions. This means that the total momentum of the objects before the collision is equal to the total momentum after the collision. In an inelastic collision, the objects may stick together or deform, but the total momentum remains the same.

5. What factors can affect the outcome of an inelastic collision?

Some factors that can affect the outcome of an inelastic collision include the masses and velocities of the objects involved, the angle and surface of the collision, and the presence of external forces such as friction or air resistance. Additionally, the type of material and elasticity of the objects can also play a role in the outcome of the collision.

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