Two bodies of masses 0.1, kg and 0.4, kg move | vedclass

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Question

According to conservation of momentum;

$m_{1} v_{1}+m_{2} v_{2}=\left(m_{1}+m_{2}\right) v$

where $\mathrm{v}$ is common velocity of the two bod

Vedclass · Accepted Answer

$1.2$

Vedclass · Answer

According to conservation of momentum;

$m_{1} v_{1}+m_{2} v_{2}=\left(m_{1}+m_{2}ight) v$

where $\mathrm{v}$ is common velocity of the two bodies

$\mathrm{m}_{1}=0.1 \mathrm{kg}, \mathrm{m}_{2}=0.4 \mathrm{kg}$

$v_{1}=1 \mathrm{m} / \mathrm{s}, \quad \mathrm{v}_{2}=-0.1 \mathrm{m} / \mathrm{s}$

$	herefore 0.1 	imes 1+0.4 	imes(-0.1)=(0.1+0.4) v$

$or\,\,0.1-0.04=0.5 v,$

$v=\frac{0.06}{0.5}=0.12 \mathrm{m} / \mathrm{s}$

Hence, distance covered $=0.12 	imes 10=1.2 \mathrm{m}$

Two bodies of masses $0.1\, kg$ and $0.4\, kg$ move towards each other with the velocities $1\, m/s$ and $0.1\, m/s$ respectively. After collision they stick together. In $10\, sec$ the combined mass travels ............... $\mathrm{m}$

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