A body of mass M1 collides elastically with another body of mass M2 at rest. After collision the velocity of M1becomes ¾ times of the velocity. Find the ratio of their masses

5 : 1

To find the ratio of the masses M1 and M2, we can use the principle of conservation of momentum. In an elastic collision, the total momentum before the collision is equal to the total momentum after the collision.

Let's assume the initial velocity of M1 is V1 and the final velocity of M1 is V1'. Additionally, let's assume the initial velocity of M2 is 0 (since it is at rest) and the final velocity of M2 is V2.

According to the principle of conservation of momentum:
Initial momentum = Final momentum

Initial momentum of the system = M1 * V1 + M2 * 0 (since the initial velocity of M2 is 0)

Final momentum of the system = M1 * V1' + M2 * V2

Setting the two equal, we get:
M1 * V1 + M2 * 0 = M1 * V1' + M2 * V2

Since we are given that after the collision the velocity of M1 becomes ¾ times of the velocity, we can write V1' = (3/4) * V1.

Substituting this value in the equation, we get:
M1 * V1 + M2 * 0 = M1 * (3/4) * V1 + M2 * V2

Simplifying the equation, we get:
M1 * V1 = (3/4) * M1 * V1 + M2 * V2

Now, we can simplify this equation further:
(4/4) * M1 * V1 = (3/4) * M1 * V1 + M2 * V2

(4M1 * V1) - (3M1 * V1) = M2 * V2 [Simplifying further]

M1 * V1 = M2 * V2

Since we know that the velocity of M1 is ¾ times the velocity, we can say V1 = (3/4) * V2.

Substituting this value in the equation, we get:
M1 * (3/4) * V2 = M2 * V2

Simplifying the equation further:
(3M1 * V2) - (4M2 * V2) = 0

Dividing the equation by V2, we get:
(3M1 - 4M2) = 0

Now, to find the ratio of the masses, we divide both sides of the equation by M2:
(3M1 - 4M2) / M2 = 0 / M2

3M1/M2 - 4 = 0

Adding 4 to both sides of the equation:
3M1/M2 = 4

Now, to find the ratio of the masses (M1/M2), we can cross-multiply:
3M1 = 4M2

Dividing both sides by M2, we get:
M1/M2 = 4/3

So, the ratio of their masses is 4/3.