Inelastic collision with an inclined plane

In summary, the conversation discusses a problem involving a particle being thrown upwards and colliding with an inclined surface. The collision is perfectly inelastic, meaning the objects stick together. The discussion includes equations and attempts to solve for the maximum height attained by the particle from the ground. It is determined that the total height above ground is 6 meters. The conversation also touches on the concept of perfectly inelastic collisions and how the particle will move after the collision.
  • #1
Vinita
23
0

Homework Statement


A particle is thrown upward with speed 10√3 m/sec. It strikes an inclined surface. Collision between the particle and inclined plane is perfectly inelastic. What will be the maximum height attained by the particle from the ground.
physics (2).jpg


2. Relevant information and equations
  • In perfectly inelastic collision (e=0) colliding objects stick to each other.
  • When net F external = 0, momentum is conserved.
  • v2 = u2 + 2as (for constant acceleration a)

The Attempt at a Solution


Just before collision let velocity of particle be v
⇒ v2 = (10√3)2 - 2g(3)
⇒ v = 4√15 m/s (g=10 m/s2

Let m = mass of particle, M = mass of surface, V = velocity of particle and surface along the normal.

Along the normal,
∫Ndt = m(V-vcos45°) = MV
⇒V = mv/√2(m-M)
Since denominator is very large, V≈ 0

Now component of velocity along the inclined plane ( = vsin45°) will not change as the normal reaction (which will give the impulse during collision) is acting to perpendicular to it.

So finally particle will move along the incline particle( upwards) till its velocity becomes zero.
0 = (vsin45°)2 - 2s(gsin45°)
⇒ s = 6√2

Thus, total height above ground = s(sin45°) + 3 = 9 m
But the answer is 6 m.
Is it because I neglected the velocity along the normal?
 

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  • #2
Vinita said:
In perfectly inelastic collision (e=0) colliding objects stick to each other.
That is only true for head on collisions.
Vinita said:
So finally particle will move along the incline
It will move that way immediately after impact, but there is no glue holding it to the ceiling.
 
  • #3
haruspex said:
That is only true for head on collisions.
But they will still move together along the normal, right? Because velocity of separation = 0
haruspex said:
It will move that way immediately after impact, but there is no glue holding it to the ceiling
So, will it be like a projectile?
I don't understand how that's possible if the surface and ball are moving together along normal. What does perfectly inelastic collision signify here?
 
  • #4
Vinita said:
But they will still move together along the normal, right?
Only at first.
Vinita said:
So, will it be like a projectile?
Yes.
Vinita said:
What does perfectly inelastic collision signify here?
That immediately after collision the velocities are the same in the direction of the impulse, so zero in this case.
Gravity continues to act, and its component in the direction of impulse is away from the ceiling, so the particle will separate from it.
 
  • #5
haruspex said:
Only at first.
Okay I get it now.
At maximum height, velocity along y-axis = 0
⇒ 0 = (v/2)2 - 2gs
⇒ s = 3 m
Total height above ground = 3 +3 = 6 m
Thanks you for helping me.:smile:
 
  • #6
Vinita said:
Okay I get it now.
At maximum height, velocity along y-axis = 0
⇒ 0 = (v/2)2 - 2gs
⇒ s = 3 m
Total height above ground = 3 +3 = 6 m
Thanks you for helping me.:smile:
You are welcome.
 
  • #7
Just to interject that you really do not need to solve the parabolic motion to solve the problem. You can just equate the difference in potential energy to the loss of the contribution of the y-component of velocity to kinetic energy.
 
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  • #8
Orodruin said:
Just to interject that you really do not need to solve the parabolic motion to solve the problem. You can just equate the difference in potential energy to the loss of the contribution of the y-component of velocity to kinetic energy.
Yeah, I didn't think of that. Thank you.
 

Related to Inelastic collision with an inclined plane

1. What is an inelastic collision with an inclined plane?

An inelastic collision with an inclined plane is a type of collision that occurs between two objects, where one of the objects is a plane that is inclined at an angle. In this type of collision, the kinetic energy of the objects is not conserved, as some of it is transferred into other forms of energy, such as heat or sound.

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

In an inelastic collision, the objects involved stick together after the collision and there is a loss of kinetic energy. In an elastic collision, the objects bounce off each other and there is no loss of kinetic energy.

3. How does the angle of inclination affect an inelastic collision with an inclined plane?

The angle of inclination can affect the outcome of the collision in several ways. A steeper angle will result in a shorter distance traveled by the objects after the collision, while a shallower angle will result in a longer distance. Additionally, a steeper angle may result in a greater loss of kinetic energy.

4. What are some real-life examples of inelastic collisions with inclined planes?

One example of an inelastic collision with an inclined plane is a car crashing into a barrier on a curved road. Another example is a ball rolling down a ramp and colliding with a stationary object at the bottom of the ramp.

5. How is the conservation of momentum applied in an inelastic collision with an inclined plane?

The conservation of momentum states that the total momentum of a closed system remains constant before and after a collision. In an inelastic collision with an inclined plane, the total momentum of the objects before the collision will be equal to the total momentum after the collision, even though some kinetic energy may have been lost.

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