Question 41: Centre of Mass and Rotational Motion

Question

A body of mass 14 kg initially at rest explodes and breaks into three fragments of masses in the ratio $2: 2: 3$. The two pieces of equal masses fly off perpendicular to each other with a speed of $18 \mathrm{~m} / \mathrm{s}$ each. The velocity of the heavier fragment is
$\_\_\_\_$ $\mathrm{m} / \mathrm{s}$.

Accepted Answer

The total mass is 14 kg , divided in the ratio 2 : 2 : 3.
Let the common multiplier be x .
$$ 2 x+2 x+3 x=14 \Rightarrow 7 x=14 \Rightarrow x=2 $$
The masses of the three fragments are :
1. $\mathrm{m}_1=2 	imes 2=4 \mathrm{~kg}$
2. $\mathrm{m}_2=2 	imes 2=4 \mathrm{~kg}$
3. $\mathrm{m}_3=3 	imes 2=6 \mathrm{~kg}$ (Heavier fragment)
The two fragments of 4 kg each fly off perpendicular to each other.
$\overrightarrow{\mathrm{p}}_1=\mathrm{m}_1 \overrightarrow{\mathrm{v}}_1=4 	imes 18 \hat{\mathrm{i}}=72 \hat{\mathrm{i}} \mathrm{kg} \cdot \mathrm{m} / \mathrm{s}$
$$\Rightarrow $$ $\overrightarrow{\mathrm{p}}_2=\mathrm{m}_2 \overrightarrow{\mathrm{v}}_2=4 	imes 18 \hat{\jmath}=72 \hat{\jmath} \mathrm{~kg} \cdot \mathrm{~m} / \mathrm{s}$
The resultant momentum of these two fragments is :
$$ \mathrm{p}_{\mathrm{res}}=\sqrt{\mathrm{p}_1^2++\mathrm{p}_2^2}=\sqrt{72^2+72^2}=72 \sqrt{2} \mathrm{~kg} \cdot \mathrm{~m} / \mathrm{s} $$
Initially, the body was at rest, so the total initial momentum is zero. Therefore, the final total momentum must also be zero:
$$ \overrightarrow{\mathrm{p}}_1+\overrightarrow{\mathrm{p}}_2+\overrightarrow{\mathrm{p}}_3=0 \Rightarrow \overrightarrow{\mathrm{p}}_3=-\left(\overrightarrow{\mathrm{p}}_1+\overrightarrow{\mathrm{p}}_2ight) $$
This means the momentum of the third (heavier) fragment must be equal in magnitude and opposite in direction to the resultant momentum of the first two.
$$ \left|\overrightarrow{\mathrm{p}}_3ight|=\mathrm{p}_{\mathrm{res}}=72 \sqrt{2} \mathrm{~kg} \cdot \mathrm{~m} / \mathrm{s} $$
We know $\mathrm{p}_3=\mathrm{m}_3 	imes \mathrm{v}_3$ :
$6 	imes v_3=72 \sqrt{2}$
$$\Rightarrow $$ $v_3=\frac{72 \sqrt{2}}{6}$
$$\Rightarrow $$ $$ \mathrm{v}_3=12 \sqrt{2} \mathrm{~m} / \mathrm{s} $$
Therefore, the velocity of the heavier fragment is $12 \sqrt{2} \mathrm{~m} / \mathrm{s}$.

Answer

$24 \sqrt{2}$

Answer

12

Answer

$12 \sqrt{2}$

Answer

$10 \sqrt{2}$

Centre of Mass and Rotational Motion

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