Question

answer the attached question​

Answer 1

The velocity of centre of mass is given by,

$\longrightarrow \bf{\bar v}=\dfrac{\sf{m_1}\bf{v_1}+\sf{m_2}\bf{v_2}+\sf{m_3}\bf{v_3}}{\sf{m_1+m_2+m_3}}$

Here $\sf{m_1+m_2+m_3=M,}$ total mass of the system.

Then,

$\longrightarrow \bf{\bar v}=\dfrac{\sf{m_1}\bf{v_1}+\sf{m_2}\bf{v_2}+\sf{m_3}\bf{v_3}}{\sf{M}}$

$\longrightarrow\sf{M}\bf{\bar v}=\sf{m_1}\bf{v_1}+\sf{m_2}\bf{v_2}+\sf{m_3}\bf{v_3}$

Here $\sf{m_1}\bf{v_1},\ \sf{m_2}\bf{v_2}$ and $\sf{m_3}\bf{v_3}$ are linear momenta of each particle. Take,

• $\bf{p_1}=\sf{m_1}\bf{v_1}$

• $\bf{p_2}=\sf{m_2}\bf{v_2}$

• $\bf{p_3}=\sf{m_3}\bf{v_3}$

Then,

$\longrightarrow\sf{M}\bf{\bar v}=\bf{p_1}+\bf{p_2}+\bf{p_3}$

In the question,

• $\bf{\bar v}=\sf{\left<30,\ 39,\ 48\right>}$

• $\bf{p_1}=\sf{\left<1,\ 2,\ 3\right>}$

• $\bf{p_2}=\sf{\left<4,\ 5,\ 6\right>}$

• $\bf{p_3}=\sf{\left<5,\ 6,\ 7\right>}$

Then,

$\longrightarrow\sf{M\left<30,\ 39,\ 48\right>=\left<1,\ 2,\ 3\right>+\left<4,\ 5,\ 6\right>+\left<5,\ 6,\ 7\right>}$

$\longrightarrow\sf{\left<30M,\ 39M,\ 48M\right>=\left<10,\ 13,\ 16\right>}$

By equating each component,

• $\sf{30M=10}$

• $\sf{39M=13}$

• $\sf{48M=16}$

we get,

$\sf{\longrightarrow\underline{\underline{M=\dfrac{1}{3}\ kg}}}$

Hence the total mass of the system is 1/3 kg.

Answer 2

Answer:

2N₂O, → 4NO₂ + O₂

(A) 2 mol of N₂O, on dissociation gives 4 mol of

NO, and 1 mol of O

(B) 1 mol of N₂O5 on dissociation gives 2 mol of

NO, and 0.5 mol of O (C) 2 g of N₂O, on dissociation gives 4 g of NO and 1 g of O₂

(D) 216 g of N₂O, on dissociation gives 184 g of

NO, and 32 g of O₂

Explanation:

2N₂O, → 4NO₂ + O₂

(A) 2 mol of N₂O, on dissociation gives 4 mol of

NO, and 1 mol of O

(B) 1 mol of N₂O5 on dissociation gives 2 mol of

NO, and 0.5 mol of O

(C) 2 g of N₂O, on dissociation gives 4 g of

NO and 1 g of O₂

(D) 216 g of N₂O, on dissociation gives 184 g of

NO, and 32 g of O₂