An 81 kg football player moving 6.5 m/s tackles and collides with a stationary 140 kg football player. What speed will the football players have in the moment after impact?

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Answer 1 · 2017-10-11T11:34:18

$4.9ms^-1$

First and foremost we measure $KE_1$ (The kinetic energy of the individual tackling):

$KE = 1/2*m*v^2$

Where;

$:. KE_1 = 1/2 * 81 * (6.5)^2$

$rarr KE_1 = 1711.125$

Due to accuracy purposes, we shall take $1711.125$ completely for $KE_1$

Now, according to the law of conservation of energy, $KE_1$ will be equal to $KE_2$ ( the kinetic energy tranferred form the tackilng player to the colliding player)

$:.$ $KE_2$ = $1/2*m*v^2$

$rarr$ $1711.125=1/2*140*v^2$

And,

$(140v^2)/2$ = $1711.125$

$140v^2$ = $2(1711.125)$

$v^2=(2(1711.125))/140$

$v=sqrt((2(1711.125))/140)$

$rarr v =4.944ms^-1$

I hope I was useful to solve your concern
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