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

Question.
An object of mass 1 kg travelling in a straight line with a velocity of 10 ms–1 collides with, and sticks to a stationary wooden block of mass 5 kg. Then they both move off together in the same straight line. Calculate the total momentum just before the impact and just after the impact. Also, calculate the velocity of the combined object.

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