a particle a of mass makes perfectly elastic collision with b which is at rest and fly apart in opp directions with same velocity then relation between their masses is
To determine the relationship between the masses of particles A and B after a perfectly elastic collision, we can use the conservation of momentum and the conservation of kinetic energy.
Given that particle B is initially at rest, its momentum (mass x velocity) is zero before and after the collision. Therefore, we only need to consider the momentum of particle A.
Let's denote the mass of particle A as mA and its initial velocity as vA. After the collision, particle A and particle B move in opposite directions with the same velocity, which we can denote as v.
Since momentum is conserved, the total momentum before the collision is equal to the total momentum after the collision.
Before the collision:
Momentum of particle A = mA * vA
After the collision:
Momentum of particle A = mA * v
Momentum of particle B = -mB * v (negative sign indicates the opposite direction)
Since momentum is conserved, we can equate the momentum before and after the collision:
mA * vA = mA * v + (-mB * v)
Simplifying the equation:
mA * vA = (mA - mB) * v
Dividing both sides of the equation by v:
mA / (mA - mB) = v / vA
Since v is the same for both particles after the collision, the ratio of their masses is given by:
mA / (mA - mB) = v / vA
Therefore, the relationship between the mass of particle A (mA) and the mass of particle B (mB) is:
mA / (mA - mB) = v / vA
Ma*Va + Mb*Vb = Ma*V - Mb*V.
Ma*Va + Mb*0 = V(Ma-Mb),
Divide both sides by V:
Ma*Va/V = Ma-Mb,
Divide both sides by Ma:
Va/V = 1-Mb/Ma,
Mb/Ma = 1 - Va/V.