A bomb of mass 8 Kg exlpodes on a smooth horizontal surface , into three fragments of masses 1 Kg , 3 Kg and 4Kg . 4Kg fragment moves with a speed 1 m/ s along eastward and 3 Kg fragement moves with a speed 1 m/s along northward . find the magnitude and direction of the velocity of 1Kg fragment .

Explanation

According to the law conservation of momentum . From the figure we get ,
along east - west direction ,1 `v cos theta =4 xx 1`
or ` v cos theta = 4 `
along north - south direction `1 * v sin theta = 3 xx1`
or ` v sin theta = 3 `
` therefore `, `v^(2) ( cos ^(2) theta + sin ^(2) theta ) = 4^(2) + 3^(2) `
or `v^(2) = 25 or , v = 5 m * s^(-1) `
Again `, ( v sin theta ) /( v cos theta ) = (3) /(4) `
` or , tan theta = ( 3) /(4) or , theta = tan ^(-1 ( 3) /(4) `
`therefore ` the velocity of the 1Kg fragement is `5 m * s^(-1) ` in a direction making an angle of ` tan ^(-1) (3)/(4) ` south odf west .