A 7.28-kilogram bowling ball traveling 8.50 meters per second east collides head-on with a 5.45 kilogram bowling ball traveling 10.0 meters per second west. Determine the magnitude of the total momentum of the two-ball system after the collision
Considering Conservation of Momentum, the momentum \(p=mv\) before and after must remain the same:
so:
before: \(7.28*8.50-5.45*10.0=7.38 kg \frac{m}{s}\)
Given:
Mass,
Speed,
By using the law of conservation of linear momentum, we get
→ \(P_{initial} = P_{final}\)
or,
→ \(P_{final } = m_1\vec{v_{1i}}+m_2\vec{v_{2i}}\)
By substituting the values, we get
\(|P_{final}|= 7.28\times 8.5-5.45\times 10\)
\(=61.88-54.5\)
\(= 7.38 \ kg.ms^{-1}\)
Thus the above approach is correct.
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