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2 years ago

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A 91.5 kg football player running east at 2.73 m/s tackles a 63.5 kg player running east at 3.09 m/s. what is their velocity aft

erward? PLEASE HELP

1 answer:

slava [35]2 years ago

6 0

Answer:

Velocity of the players afterwards = 2.88 m/s towards east.

Explanation:

Mass of football player A $(m_1)$ = 91.5 kg

Velocity of player A $(v_1)$ = 2.73 m/s

Mass of football player B $(m_2)$ =63.5 kg

Velocity of player B $(v_2)$ = 3.09 m/s

Since both players move in same direction east, so their velocity afterwards will also be in same direction east.

By law of conservation of momentum we have:

Momentum before collision = Momentum after collision

This can be written as:

$p_i=p_f$

$m_1v_1+m_2v_2=(m_1+m_2)\ v$

where $v$ is the velocity of the players together afterwards.

We can plugin the given value to find $v$

$(91.5)(2.73)+(63.5)(3.09)=(91.5+63.5)\ v$

$249.795+196.215=155\ v$

$446.01=155\ v$

Dividing both sides by 155.

$\frac{446.01}{155}=\frac{155\ v}{155}$

$2.88=v$

∴ $v=2.88$ m/s

Velocity of the players afterwards = 2.88 m/s towards east.

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