A 91.5 kg football player running east at 3.73 m/s tackles a 63.5 kg player running east at 3.09 m/s. what is their velocity afterward? PLEASE HELP

Respuesta :

Answer:

Their velocity afterwards = [tex]3.47\ ms^{-1}[/tex]

Explanation:

Mass of Player 1 = [tex]m_1[/tex]= [tex]91.5\ kg[/tex]

Mass of Player 2 =  [tex]m_2[/tex]= [tex]63.5\ kg[/tex]

Velocity of Player 1 moving east =[tex]3.73\ ms^{-1}[/tex]

Velocity of Player 2 moving left =[tex]3.09\ ms^{-1}[/tex]

By law of conservation of momentum:

[tex]m_1v_1+m_2v_2=(m_1+m_2)v[/tex]

Here [tex]v[/tex] represents the velocity of tangled players.

Plugging in values in the equation.

[tex](91.5\times 3.73)+(63.5\times 3.09)=(91.5+63.5)v[/tex]

[tex]341.295+196.215=(155)\times v[/tex]

[tex]537.51=(155)\times v[/tex]

Dividing both sides by 155.

[tex]\frac{537.51}{155}=\frac{(155)\times v}{155}[/tex]

∴ [tex]v=3.467\approx 3.47\ ms^{-1}[/tex]

Their velocity afterwards = [tex]3.47\ ms^{-1}[/tex]

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