Question

At the top of trajectory ,a mortar shell explodes into two fragments ;a 2.5kg pieces moves south west with a horizontal velocity of 100m/s while 3.5kg piece moves north with a horizontal velocity of 70 m/s.what was the horizontal direction of the cell just before explosion

Answer 1

$m_1=2.5 \; kg$

$m_2=3.5 \; kg$

$m=m_1+m_2=6\; kg$

$v_1=100 \;m/s$

$v_2=70\; m/s$

$v-?$

Solution:

Total momentum before the explosion is equal to the total momentum after the explosion:

$p=p_1+p_2$

$p=m\overline{v}, \; p_1=m_1\overline{v_1}, p_2=m_2\overline{v_2}$

$6\overline{v}=2.5\overline{v_1}+3.5\overline{v_2}$

x-axis:

$6v_x=2.5v_{1x}+3.5v_{2x}=2.5\times (-\cos(45^{\circ}))\times 100+3.5\times\cos(90^{\circ})\times70=$

$=2.5\times(-\frac{1}{\sqrt{2}})\times 100+0= -125\sqrt{2}$

$v_x=\frac{-125\sqrt{2}}{6}$

y-axis:

$6v_y=2.5v_{1y}+3.5v_{2y}=2.5\times (-\cos(45^{\circ}))\times 100+3.5\times\cos(0^{\circ})\times70=$

$=2.5\times(-\frac{1}{\sqrt{2}})\times 100+3.5\times1\times70=245 -125\sqrt{2}$

$v_y=\frac{245-125\sqrt{2}}{6}$

$v=\sqrt{{v_1}^2+{v_2}^2}=\frac{1}{6}\sqrt{(-125\sqrt{2})^2+(245-125\sqrt{2})^2}=31.58 \;m/s$

Answer: $v=31.58 \; m/s.$