Question

A bomb initially at rest explodes by itself into three equal masses fragments the velocity of two fragments are

Answer 1

Answer:

Complete the question

Answer 2

Correct Question: A bomb initially at rest explodes by itself into three equal masses fragments the velocity of two fragments are (3i + 2j) m/s and (-i - 4j) m/s . The velocity of the third fragment is?

Answer:

The velocity of the third fragment is (-2i + 2j) ms⁻¹

Explanation:

Given;-

    Masses are, = same for all fragments.

Velocities are, V₁ =  (3i + 2j) m/s

                        V₂ = (-i - 4j) m/s

So, let the velocity of the third fragment be V₃.

Let Vxcm represent velocity of Vcm in x-axis & Vycm represent velocity of Vcm in y-axis.

Now, we know by formula that;-

          Vxcm = Vx₁m + Vx₂m + Vx₃m / m + m + m

                 0 = 3i(m) - i(m) + Vx₃(m) / 3m

                 0 = m [ 3i - i + Vx₃]/ 3m

                 0 = 2i + Vx₃

              Vx₃ = - 2i _______(1)

Similarly;-

          Vycm = Vy₁m + Vy₂m + Vy₃m / m + m + m

                 0 = 2j(m) - 4j(m) + Vy₃(m) / 3m

                 0 = m[ 2j - 4j + Vy₃] / 3m

                 0 = - 2j + Vy₃

              Vy₃ = 2j _________(2)

From (1) and (2), we obtain the velocity of the third fragment as;-

           Vcm = (-2i + 2j) ms⁻¹

Hence, the velocity of the third fragment is (-2i + 2j) ms⁻¹

Hope it helps! ;-))