Conservation of Momentum EX 5

The mass of the third piece is (6 - 3 - 2)kg = 1 kg. We will compute the components of the linear momentum of the bomb, $p_b$, the 1-kg piece, $p_1$, the 2-kg piece, $p_2$, and the 3-kg piece, $p_3$. The linear momentum is conserved. We predict that the vector $p_3$ must be in the 4th quadrant so that $p_b=p_1+p_2+p_3$.

Momentum	x-component	y-component
$p_b$	$(6)(5)\cos (-37^{\circ })=24\frac{kgm}{s}$	$(6)(5)\sin (-37^{\circ })=18\frac{kgm}{s}$
$p_1$	${p_1}_x$	${p_1}_y$
$p_2$	$-(2)(3)=-6\frac{kgm}{s}$	0
$p_3$	$(3)(2)\cos (53^{\circ })=3.6\frac{kgm}{s}$	$(3)(2)\sin (53^{\circ })=4.8\frac{kgm}{s}$

Then we write the equations

$p_{{before}_x}=p_{{after}_x}$

$24={p_1}_x-6+3.6$

${p_1}_x=26.4\frac{kgm}{s}$

and

$p_{{before}_y}=p_{{after}_y}$

$-18={p_1}_y+0+4.8$

${p_1}_y=-22.8\frac{kgm}{s}$

To obtain the velocity of the third fragment, we divide its momentum by its mass (1 kg):

$\overrightarrow{v}=26.4\hat{i}-22.8\hat{j}$

We note that our prediction was right. The linear momentum of the third piece is in the fourth quadrant.