Question 41: Centre of Mass and Rotational Motion

Question

Two blocks of masses 2 kg and 1 kg respectively, are tied to the ends of a string which passes over a light frictionless pulley as shown in the figure below. The masses are held at rest at the same horizontal level and then released. The distance traversed by the centre of mass in 2 s is _______ m. (Take $g = 10 \; m/s^2$)

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

3.33

Answer

3.12

Answer

2.22

Answer

1.42

Centre of Mass and Rotational Motion

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