Question

A trolley of mass 4kg moving on a smooth horizontal platform with a speed of 1.0m/s collide perfectly mass on the same platform. Calculate the total movement of the two trolleys immediately after collision.

Answer 1

In a perfectly inelastic collision, the two objects stick together after colliding. To find the total momentum before and after the collision, you can use the principle of conservation of momentum.

1. **Calculate the initial momentum**:

Before the collision, only the moving trolley has momentum:
- Mass of trolley 1 (m₁) = 4 kg
- Velocity of trolley 1 (v₁) = 1.0 m/s
- Mass of trolley 2 (m₂) = 0 kg (as it’s initially at rest)
- Velocity of trolley 2 (v₂) = 0 m/s

Initial momentum (p_initial):
\[
p_{\text{initial}} = m_1 \cdot v_1 + m_2 \cdot v_2 = 4\, \text{kg} \cdot 1.0\, \text{m/s} + 0\, \text{kg} \cdot 0 = 4\, \text{kg m/s}
\]

2. **Calculate the final velocity after collision**:

Since the trolleys stick together after the collision:
- Total mass after collision (M) = m₁ + m₂ = 4 kg + 0 kg = 4 kg

According to the conservation of momentum, the total momentum before the collision equals the total momentum after the collision:
\[
p_{\text{initial}} = p_{\text{final}}
\]
Let \( V_f \) be the final velocity of the two trolleys combined after the collision:
\[
p_{\text{final}} = (m_1 + m_2) \cdot V_f = 4\, \text{kg} \cdot V_f
\]

Setting the initial and final momentum equal:
\[
4\, \text{kg m/s} = 4\, \text{kg} \cdot V_f
\]

Now, dividing both sides by 4 kg:
\[
V_f = \frac{4\, \text{kg m/s}}{4\, \text{kg}} = 1.0\, \text{m/s}
\]

3. **Conclusion**:
The total movement (momentum) of the two trolleys immediately after the collision remains the same as before:
- The final velocity of the two trolleys together is 1.0 m/s.
- The total momentum after the collision is also 4 kg m/s.