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4-1.Newton's Laws of Motion
normal
A bomb is projected with $200\,m/s$ at an angle $60^o$ with horizontal. At the highest point, it explodes into three particles of equal masses. One goes vertically upward with velocity $100\,m/sec,$ second particle goes vertically downward with the same velocity as the first. Then what is the velocity of the third one
A$120\,m/sec$ with $60^o$ angle
B$200\,m/sec$ with $30^o$ angle
C$50\,m/sec,$ in horizontal direction
D$300\,m/sec,$ in horizontal direction
Solution
By $COLM$ in $Hz\, dir^n$
$\mathrm{m}(100)=\frac{\mathrm{m}}{3} \mathrm{v}^{\prime}$
$\mathrm{V}^{\prime}=300 \mathrm{m} / \mathrm{s}$ in $\mathrm{Hz}\, \mathrm{dir}^{n}$
$\mathrm{m}(100)=\frac{\mathrm{m}}{3} \mathrm{v}^{\prime}$
$\mathrm{V}^{\prime}=300 \mathrm{m} / \mathrm{s}$ in $\mathrm{Hz}\, \mathrm{dir}^{n}$
Standard 11
Physics
