Physics: A91.5 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?

A91.5 kg football player running east at 3.73 m/s - id9153733799019, wdymRachel4943

tleppek6245

10.01.2022

Their velocity afterwards = $3.47\ ms^{-1}$

Explanation:

Mass of Player 1 = m_1 = $91.5\ kg$

Mass of Player 2 = m_2 = $63.5\ kg$

Velocity of Player 1 moving east = $3.73\ ms^{-1}$

Velocity of Player 2 moving left = $3.09\ ms^{-1}$

By law of conservation of momentum:

m_1v_1+m_2v_2=(m_1+m_2)v

Here represents the velocity of tangled players.

Plugging in values in the equation.

$(91.5\times 3.73)+(63.5\times 3.09)=(91.5+63.5)v$

$341.295+196.215=(155)\times v$

$537.51=(155)\times v$

Dividing both sides by 155.

$\frac{537.51}{155}=\frac{(155)\times v}{155}$

∴ $v=3.467\approx 3.47\ ms^{-1}$

Their velocity afterwards = $3.47\ ms^{-1}$

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