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5.Work, Energy, Power and Collision
normal
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 ............... $\mathrm{m}$
A
$120$
B
$0.12$
C
$12$
D
$1.2$
Solution
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 bodies
$\mathrm{m}_{1}=0.1 \mathrm{kg}, \mathrm{m}_{2}=0.4 \mathrm{kg}$
$v_{1}=1 \mathrm{m} / \mathrm{s}, \quad \mathrm{v}_{2}=-0.1 \mathrm{m} / \mathrm{s}$
$\therefore 0.1 \times 1+0.4 \times(-0.1)=(0.1+0.4) v$
$or\,\,0.1-0.04=0.5 v,$
$v=\frac{0.06}{0.5}=0.12 \mathrm{m} / \mathrm{s}$
Hence, distance covered $=0.12 \times 10=1.2 \mathrm{m}$
Standard 11
Physics
