An object of mass 1 kg travelling in a straight line with a velocity of 10 ms–1 collides with,

Solution:

For object : $\mathrm{m}_{1}=1 \mathrm{~kg} ; \mathrm{u}_{1}=10 \mathrm{~ms}^{-1}$

For wooden block: $\mathrm{m}_{2}=5 \mathrm{~kg} ; \mathrm{u}_{2}=0$

Momentum just before collision

$=\mathrm{m}_{1} \mathrm{u}_{1}+\mathrm{m}_{2} \mathrm{u}_{2}=1 \times 10+5 \times 0=10 \mathrm{~kg}

\mathrm{~ms}^{-1}$

Since, momentum is conserved,

momentum before collision $=$ momentum after collision $=10 \mathrm{~kg} \mathrm{~ms}^{-1}$

Mass after collision $=\left(\mathrm{m}_{1}+\mathrm{m}_{2}\right)=1+5=6 \mathrm{~kg}$

Let velocity after collision $=\mathrm{v}$

$\therefore$ Momentum after collision $=6 \times \mathrm{v}$

Using the law of conservation of momentum,
momentum after collision = momentum before collision

$\therefore 6 \times \mathrm{v}=10$ or $\mathrm{v}=\frac{10}{6}=1.67 \mathrm{~ms}^{-1}$