Estimate the velocity of each stone have just after the collision

In summary, the conversation discusses a collision between two curling stones, A and B, on an ice rink. Stone A's velocity before the collision is 2m/s and after the collision it moves at a 45 degree angle at 1.5m/s. Stone B's velocity after the collision is unknown but can be calculated using the conservation of momentum. The collision is assumed to be approximately elastic and the loss of kinetic energy is attributed to ice-sludge between the stones. More information is requested but it is suggested to use the formula and principle of conservation of momentum to find the answer. The conversation also includes a resource for understanding momentum and acknowledges that the person asking for information is not a native English speaker.
  • #1
Newton86
59
0
Hi :smile: I really need a answer to this task. Hope anyone can help.

A curling stone ( named A ) gets on the ice with a unknown beginning velocity Vo
the stone moves along a straight line ( x-axis ) on the x-axis further ahead lies another
stone ( named B ) still. Stone A collides with stone B in a sloped hit/knock
Just before the hit stone A has a velocity of 2m/s. After the slope stone A moves
straight with a angle of 45 degree in proportion at/of the x axis. Equivalently for stone B is -30 degrees.

a ) Estimate the velocity of each stone have just after the collision

b) Estimate the loss of kinetic energy in this collision. A curling stone have mass 20kg. what kind of collision is this ?


The reason for the energy loss is that big is that ice-sludge came between the stones in the collision. We clean up and retake the attempt. Now we reckon that the push is approximate perfect elastic Stone A has the velocity of 2 m/s before the push/hit and 1,5m/s after. The direction of a form the angle 41,3 degree with the x-axis after the push/hit.

c) Find the velocity and direction of B after the stone is Pushed/released.

d ) Check that the push of the stone is approximated elastic, by calculate the x component
 
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  • #2
a) use conservation of momentum in the x and in the y direction.
 
  • #3
Thanks. But can I get some moore information I am a noob :)
 
  • #4
none ?
 
  • #5
Newton86 said:
Thanks. But can I get some moore information I am a noob :)
What further information do your require?
 
  • #6
Hootenanny said:
What further information do your require?

The formula and correct answer would been nice
 
  • #7
Newton86 said:
The formula and correct answer would been nice
Nope, we don't do that here. Your going to have to do some work yourself. What can you say about the total momentum before and after the collision?
 
  • #8
Hootenanny said:
Nope, we don't do that here. Your going to have to do some work yourself. What can you say about the total momentum before and after the collision?

Its less after the collition (velocity) ? Not sure what mometum means movement/direction or velocity ? But I am guessing its velocity
Im Not english:blushing:
 
  • #9
Heres a good elementary resource: http://www.glenbrook.k12.il.us/GBSSCI/PHYS/CLASS/momentum/u4l1a.html
 
Last edited by a moderator:
  • #10
But in a you don't know the mass
 

1. What is the formula for calculating the velocity of a stone after a collision?

The formula for calculating velocity after a collision is v = (m1u1 + m2u2)/(m1 + m2), where v is the final velocity, m1 and m2 are the masses of the colliding objects, and u1 and u2 are the initial velocities of the objects.

2. How is the velocity of each stone determined after a collision?

The velocity of each stone is determined by using the law of conservation of momentum, which states that the total momentum before a collision is equal to the total momentum after the collision. This means that the sum of the momentums of the two stones before the collision is equal to the sum of the momentums after the collision.

3. Is the velocity of each stone the same after a collision?

No, the velocity of each stone after a collision can vary depending on the masses and initial velocities of the objects involved. In an elastic collision, the velocities of the objects after the collision can be calculated using the formula v1 = (m1u1 - m2u2)/(m1 + m2) and v2 = (2m1u1 + m2u2)/(m1 + m2), where v1 and v2 are the final velocities of the two stones.

4. How does the angle of collision affect the velocity of the stones?

The angle of collision can affect the velocity of the stones by changing the direction of their velocities. In an oblique collision, the final velocities of the stones can be calculated using the equations v1 = (m1u1cosθ1 + m2u2cosθ2)/(m1 + m2) and v2 = (m1u1sinθ1 + m2u2sinθ2)/(m1 + m2), where θ1 and θ2 are the angles of the initial velocities of the objects.

5. Are there any external factors that can affect the velocity of the stones after a collision?

Yes, external factors such as friction, air resistance, and other forces can affect the velocity of the stones after a collision. In real-world situations, it is important to consider these factors when estimating the velocity of the stones after a collision.

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