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Physics Question on Motion in a plane

A projectile is moving at 20 m s−120\,m\,s^{-1} at its highest point, where it breaks into two equal parts due to an internal explosion. One part moves vertically up at 30 m s−130\,m\,s^{-1} with respect to the ground. Then the other part will move at

A

1013m s−110\sqrt{13}m\,s^{-1}

B

20 m s−120\,m\,s^{-1}

C

30 m s−130\,m\,s^{-1}

D

50 m s−150\,m\,s^{-1}

Answer

50 m s−150\,m\,s^{-1}

Explanation

Solution

Using momentum conservation,
m×20i^=m/2×30j^+m/2×x x=40i^−30j^ ⇒∣x∣=50 \begin{array}{l} m \times 20 \hat{ i }= m / 2 \times 30 \hat{ j }+ m / 2 \times x \\\ x =40 \hat{ i }-30 \hat{ j } \\\ \Rightarrow| x |=50 \end{array}