To solve this problem, we can apply the principle of conservation of momentum. According to this principle, the total momentum before the collision is equal to the total momentum after the collision.
Before the collision, the momentum of particle A is given by P1 = M1 * V1, and the momentum of particle B is given by P2 = M2 * V2.
After the collision, the particle A's momentum changes to P1' = M1' * V1' (where V1' is the velocity of A after the collision), and the particle B's momentum changes to P2' = M2' * V2' (where V2' is the velocity of B after the collision).
Since the collision is perfectly elastic, the total kinetic energy before and after the collision should be equal. Therefore, we can write the following equation:
(1/2) * M1 * (V1)^2 + (1/2) * M2 * (V2)^2 = (1/2) * M1' * (V1')^2 + (1/2) * M2' * (V2')^2
Expanding this equation, we get:
(1/2) * M1 * (V1)^2 + (1/2) * M2 * (V2)^2 = (1/2) * M1' * (V1')^2 + (1/2) * M2' * (V2')^2
Multiplying each term by 2 to simplify the equation, we have:
M1 * (V1)^2 + M2 * (V2)^2 = M1' * (V1')^2 + M2' * (V2')^2
Now, using the principle of conservation of momentum, we know that the total momentum before the collision is equal to the total momentum after the collision. Therefore, we can write the following equation:
P1 + P2 = P1' + P2'
M1 * V1 + M2 * V2 = M1' * V1' + M2' * V2'
Since the collision is perfectly elastic, the velocities of the particles after the collision can be written as:
V1' = ((M1 - M2) / (M1 + M2)) * V1 + (2 * M2 / (M1 + M2)) * V2
V2' = ((2 * M1) / (M1 + M2)) * V1 + ((M2 - M1) / (M1 + M2)) * V2
Substituting these expressions into the conservation of momentum equation, we get:
M1 * V1 + M2 * V2 = ((M1 - M2) / (M1 + M2)) * V1 + (2 * M2 / (M1 + M2)) * V2 + ((2 * M1) / (M1 + M2)) * V1 + ((M2 - M1) / (M1 + M2)) * V2
Simplifying this equation, we obtain:
M1 * V1 + M2 * V2 = ((M1 - M2) * V1 + 2 * M2 * V2 + 2 * M1 * V1 + (M2 - M1) * V2) / (M1 + M2)
Multiplying each term by (M1 + M2) and simplifying further, we get:
M1 * V1 + M2 * V2 = M1 * V1 - M2 * V1 + 2 * M2 * V2 + 2 * M1 * V1 - M1 * V2 + M2 * V2
Canceling out like terms, we have:
M2 * V2 = -M2 * V1 + 2 * M1 * V1 - M1 * V2 + M2 * V2
Rearranging the terms, we obtain:
2 * M2 * V2 + M1 * V2 = M2 * V1 + 2 * M1 * V1
Factoring out V2 and V1, we have:
V2 * (2 * M2 + M1) = V1 * (M2 + 2 * M1)
Dividing both sides by (2 * M2 + M1), we obtain the final formula for V1:
V1 = (M2 - M1) / (M2 + M1) * V2 + 2 * M2 / (M2 + M1) * V1
A particle (A) of mass M1 travelling with velocity of V1 collide perfectly with a particle (B) of mass M2 travelling with velocity V2 (less than V1) show that the velocity V1 of (A) after the impact is given by,
V1=(M1- M2/M1+M2)U1+2M2 V2/M1+M2
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