A plate drops onto a smooth floor and shatters into three pieces of equal mass. Two of the pieces go off with equal speeds v at right angles to one another. Find the speedand direction of the third piece.

The plates are assumed to travel along the floor surface, such their initial

velocity along the floor plane is (zero) is used for the calculation

Reasons:

Let, m represent the mas of the plate pieces.

Let the direction of piece 1 be along the x-axis, and the direction of piece 2 be  along the y-axis, we have;

= v·i

= v·j

Let v₃ represent the speed of piece 3

Taking the initial speed of the plate as v₁, by conservation of linear momentum, we have;

(m + m + m)·v₁ = v·i·m +  v·j·m + v₃·m

Initial velocity along the horizontal x and y direction is zero, therefore;

(m + m + m)·v₁ = 0 =  v·i·m +  v·j·m + v₃·m

v₃·m = -(v·i·m +  v·j·m)

v₃ = -(v·i +  v·j)

The magnitude of the velocity of the third piece, |v₃| = √((-v)² + (-v)²) = √(2·v²)

Given that the direction of the third piece is in the negative x and y directions, we have;

Learn more about the conservation of momentum principle here:

link

Speed of the this part is given as

Also the direction of the velocity of the third part of plate is moving along 135 degree with respect to one part of the moving plate

Explanation:

As we know by the momentum conservation of the system

we will have

here we know that

the momentum of two parts are equal in magnitude but perpendicular to each other

so we will have

now from above equation we have

Also the direction of the velocity of the third part of plate is moving along 135 degree with respect to one part of the moving plate

no. dot product   =0 when vecs at right angles

i think the answer would be reserve.