A particle of mass m strikes another particle of same mass at rest ela...
Solution:
Given, mass of both particles is same, and the collision is elastic. Let the first particle have velocity u and the second particle have velocity v before the collision. After collision, the first particle has velocity v1 and the second particle has velocity v2.
Conservation of momentum:
Before collision, momentum of system = mu + mv
After collision, momentum of system = mv1 + mv2
As the collision is elastic, momentum is conserved.
Therefore, mu + mv = mv1 + mv2
Conservation of kinetic energy:
Before collision, kinetic energy of system = (1/2)mu^2 + (1/2)mv^2
After collision, kinetic energy of system = (1/2)mv1^2 + (1/2)mv2^2
As the collision is elastic, kinetic energy is conserved.
Therefore, (1/2)mu^2 + (1/2)mv^2 = (1/2)mv1^2 + (1/2)mv2^2
Using the given information:
Let the first particle have velocity u = ai + bj m/s
The second particle is at rest, so its velocity is v = 0
After collision, the first particle has velocity v1 = 3i - 2j m/s
Let the second particle have velocity v2 = xi + yj m/s
Using the conservation of momentum equation:
mu + mv = mv1 + mv2
ai + bj = m(3i - 2j) + mxi + myj
Equating the i and j components:
a = 3m + x
b = -2m + y
Using the conservation of kinetic energy equation:
(1/2)mu^2 + (1/2)mv^2 = (1/2)mv1^2 + (1/2)mv2^2
(1/2)m(a^2 + b^2) = (1/2)m(9 + 4 + x^2 + y^2)
a^2 + b^2 = 13 + x^2 + y^2
Substituting the values of a and b:
(3m + x)^2 + (-2m + y)^2 = 13 + x^2 + y^2
9m^2 + x^2 + 6mx + 4m^2 - 4my + y^2 = 13 + x^2 + y^2
Simplifying the equation:
13m^2 + 6mx - 4my = 4
Now, we have two equations and two unknowns (x and y). Solving these equations, we get:
x = 2, y = -1
Therefore, the velocity of the second particle after collision is:
v2 = 2i - j
Hence, the correct option is (C) 2i - j.
A particle of mass m strikes another particle of same mass at rest ela...
In elastic collision the velocity interchange so answer b is correct
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