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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

Answer :

Giovi
Considering Conservation of Momentum, the momentum [tex]p=mv[/tex] before and after must remain the same:
so:
before: [tex]7.28*8.50-5.45*10.0=7.38 kg \frac{m}{s} [/tex]
Cricetus

The magnitude of the total momentum will be "7.38 kg.ms⁻¹".

Given:

Mass,

  • [tex]m_1 = 7.28 \ kg[/tex]
  • [tex]m_2 = 5.45 \ mg[/tex]

Speed,

  • [tex]v_1 = 8.5 \ m/s[/tex]
  • [tex]v_2 = 10 \ m/s[/tex]

By using the law of conservation of linear momentum, we get

→ [tex]P_{initial} = P_{final}[/tex]

or,

→ [tex]P_{final } = m_1\vec{v_{1i}}+m_2\vec{v_{2i}}[/tex]

By substituting the values, we get

 [tex]|P_{final}|= 7.28\times 8.5-5.45\times 10[/tex]

             [tex]=61.88-54.5[/tex]

             [tex]= 7.38 \ kg.ms^{-1}[/tex]

Thus the above approach is correct.

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