A sphere of mass 9 kg is moving at 7 m/s to the right until it smacks into a second stationary sphere of mass 2 kg. After the collision, both spheres travel to the right: the first sphere at 4.45 m/s, and the second sphere at 11.5 m/s. What kind of collision took place?
To determine the type of collision that took place, we need to consider the law of conservation of momentum. In an isolated system, the total momentum before the collision should be equal to the total momentum after the collision.
Before the collision, the momentum of the first sphere is given by:
\[P_1 = m_1 \cdot v_1 = (9 \, \text{kg})(7 \, \text{m/s}) = 63 \, \text{Ns}\]
After the collision, the momentum of the first sphere is given by:
\[P_1' = m_1 \cdot v_1' = (9 \, \text{kg})(4.45 \, \text{m/s}) = 40.05 \, \text{Ns}\]
Similarly, for the second sphere:
Before the collision: \(P_2 = m_2 \cdot v_2 = (2 \, \text{kg})(0 \, \text{m/s}) = 0 \, \text{Ns}\)
After the collision: \(P_2' = m_2 \cdot v_2' = (2 \, \text{kg})(11.5 \, \text{m/s}) = 23 \, \text{Ns}\)
The total momentum before the collision is:
\[P_{\text{total before}} = P_1 + P_2 = 63 \, \text{Ns} + 0 \, \text{Ns} = 63 \, \text{Ns}\]
The total momentum after the collision is:
\[P_{\text{total after}} = P_1' + P_2' = 40.05 \, \text{Ns} + 23 \, \text{Ns} = 63.05 \, \text{Ns}\]
Since \(P_{\text{total before}} \neq P_{\text{total after}}\), the momentum is not conserved and the collision is inelastic.