A stationary bomb of 10kg mass explodes into 3 fragments. two of these parts having mass 4kg and 2kg,fly apart perpendicular to each other with a velocity of 2ms-1 and 3ms-1.find magnitude and direction of 3rd part.

To find the magnitude and direction of the velocity of the third fragment, we can use the law of conservation of momentum.

The law of conservation of momentum states that the total momentum of an isolated system remains constant if no external forces act upon it.

In this case, the initial momentum of the bomb is zero because it is stationary. After the explosion, the total momentum of all the fragments must still be zero.

Let's denote the velocity of the third fragment by v.

Since the first two fragments fly apart perpendicular to each other, we can treat them as independent motions.

First fragment:
Mass (m1) = 4 kg
Velocity (v1) = 2 m/s

Second fragment:
Mass (m2) = 2 kg
Velocity (v2) = 3 m/s

Since the momentum is conserved, we can write the equation:

(m1 * v1) + (m2 * v2) + (m3 * v) = 0

Plugging in the values we have:
(4 kg * 2 m/s) + (2 kg * 3 m/s) + (m3 * v) = 0

Simplifying this equation:
8 kg m/s + 6 kg m/s + (m3 * v) = 0

Rearranging the equation to solve for m3 * v:
m3 * v = -14 kg m/s

Since we don't have the individual mass or velocity of the third fragment, we can express it as the product m3 * v.

To find the magnitude and direction of the third part's velocity, we need more information.