Answer:
Explanation:
mass, m = 1.2 kg
initial velocity, u = 0.5 m/s along X axis
[tex]\overrightarrow{u}=0.5\widehat{i}[/tex]
mass of first fragment, m1 = 0.4 kg
final velocity of the first fragment, v1 = 0.9 m/s along - Z axis
[tex]\overrightarrow{v_{1}}=-0.9\widehat{k}[/tex]
mass of second fragment, m2 = 1.2 - 0.4 = 0.8 kg
Let the final velocity of the second fragment is v2.
Use conservation of momentum
[tex]m\times \overrightarrow{u}=m_{1}\times \overrightarrow{v_{1}} + m_{2}\times \overrightarrow{v_{2}}[/tex]
[tex]1.2 \times 0.5 \widehat{i}=0.4 \times (-0.9\widehat{k})+ 0.8\times \overrightarrow{v_{2}}[/tex]
[tex]\overrightarrow{v_{2}}=0.75\widehat{i}+0.45\widehat{j}[/tex]
Angle made with X axis is 0 degree.
