Question

An object of mass 1 kg travelling in a straight line with a velocity of 10 m s^–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.

Answer 1

Given:

Mass of moving object, \( m_1 = 1 \, \text{kg} \)
Velocity of moving object, \( u_1 = 10 \, \text{m/s} \)
Mass of stationary block, \( m_2 = 5 \, \text{kg} \)
Velocity of stationary block, \( u_2 = 0 \, \text{m/s} \)

Step 1: Total momentum just before impact

\[ p_{\text{before}} = m_1 u_1 + m_2 u_2 \]
\[ p_{\text{before}} = (1 \times 10) + (5 \times 0) = 10 \, \text{kgm/s} \]

Step 2: Total momentum just after impact

By the law of conservation of momentum:
\[ p_{\text{after}} = p_{\text{before}} = 10 \, \text{kgm/s} \]

Step 3: Velocity of the combined object

Let the combined mass be \( M = m_1 + m_2 = 1 + 5 = 6 \, \text{kg} \)
Velocity of combined object, \( v = \frac{p_{\text{after}}}{M} = \frac{10}{6} \approx 1.67 \, \text{m/s} \)

Answer:

Total momentum just before impact = 10 kg·m/s
Total momentum just after impact = 10 kg·m/s
Velocity of combined object = 1.67 m/s