Question

A body of mass M kg initially at rest explodes into three fragments having masses in the ratio $ 3:1:1 $ . The two equal mass fragments fly off at right angles to each other with equal speed of 60 m/s. The speed of the heavier fragment is

• A. $ 30\sqrt{2}\,m/s $
• B. $ 10\sqrt{2}\,m/s $
• C. $ 20\sqrt{2}\,m/s $
• D. $ 20\,m/s $

Answer 1

The Correct Option is C

: Momentum is conserved $ \therefore $ (3mv) = Momentum of heavy particle $ \therefore $ $ mu= $ momentum of lighter particle $ {{(mu)}^{2}}+{{(mu)}^{2}}={{(3mv)}^{2}} $ or $ 2{{m}^{2}}{{u}^{2}}=9{{m}^{2}}{{v}^{2}}\Rightarrow 2\times {{(60)}^{2}}=9{{r}^{2}} $ or $ {{v}^{2}}=\frac{2\times 60\times 60}{9}=2\times 400 $ or $ v=20\sqrt{2}m/s $