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Ball 1 collides with an another identical ball 2 at rest .for what value of coefficient of restitution e,the velocity of second ball becomes two times that of 1 after collision?
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Understanding the Collision
In a one-dimensional elastic collision between two identical balls, we can analyze the situation using the principles of conservation of momentum and the coefficient of restitution.
Initial Conditions
- Ball 1 is moving with velocity u.
- Ball 2 is at rest (velocity = 0).
Post-Collision Velocities
Let:
- v1 = final velocity of Ball 1 after the collision.
- v2 = final velocity of Ball 2 after the collision.
We need to find the coefficient of restitution (e) such that v2 = 2v1.
Conservation of Momentum
The law of conservation of momentum states that:
- Initial momentum = Final momentum
- mu + 0 = mv1 + mv2
- u = v1 + v2
Coefficient of Restitution
The coefficient of restitution is defined as:
- e = (relative velocity after collision) / (relative velocity before collision)
For our case:
- e = (v2 - v1) / (u - 0)
- e = (v2 - v1) / u
Setting Up the Equations
From the condition v2 = 2v1:
1. Substitute v2 into the momentum equation:
u = v1 + 2v1
u = 3v1
Therefore, v1 = u/3.
2. Now substitute v1 back into the expression for v2:
v2 = 2(v1) = 2(u/3) = 2u/3.
Finding the Coefficient of Restitution (e)
Now substituting these values into the coefficient of restitution equation:
- e = (v2 - v1) / u
- e = [(2u/3) - (u/3)] / u
- e = (u/3) / u = 1/3.
Conclusion
The required coefficient of restitution for the velocity of the second ball to become two times that of the first ball after collision is e = 1/3.
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Ball 1 collides with an another identical ball 2 at rest .for what value of coefficient of restitution e,the velocity of second ball becomes two times that of 1 after collision?
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