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.

Given, mass of the object \( m_1 \) = 1 kg

Mass of the block \( m_2 \) = 5 kg

Initial velocity of the object \( u_1 \) = 10 m/s

Initial velocity of the block \( u_2 \) = 0

Mass of the resulting object \( = m_1 + m_2 = 6 kg \)

Total momentum before the collision = \( m_1 u_1 + m_2 u_2 = (1) × (10) + 0 = 10 kg m/s \)

By law of conservation of momentum, the total momentum before the collision is equal to the total momentum post the collision.

\(\therefore \) The total momentum before the collision is also 10 kg m/s

Now, \( (m_1 + m_2) × v = 10 kg m/s \)

\( \Rightarrow v = \frac{10}{6} = 1.66 m/s \)