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At the county fair, Chris throws a 0.15 kg baseball at a 2.0 kg wooden milk bottle, hoping to knock it off its stand and win a prize. The ball bounces straight back at 20% of its incoming speed, knocking the bottle straight forward. What is the bottle’s speed, as a percentage of the ball’s incoming speed?

Respuesta :

Answer:

So percentage of the incoming speed is 9% by which bottle will move

Explanation:

As we know that ball is projected directly towards the bottle

So here momentum given by the ball is same as momentum gain by the bottle

So first we will find the change in momentum of the ball

[tex]\Delta P_{ball} = m(v_f - v_i)[/tex]

Here we know that

[tex]v_f = 0.20 v_i[/tex]

[tex]\Delta P_{ball} = 0.15( 0.20 v - (-v))[/tex]

[tex]\Delta P_{ball} = 0.18 v[/tex]

Now momentum gain by the bottle is

[tex]0.18 v = 2 v_b[/tex]

[tex]v_b = 0.09 v[/tex]

So percentage of the incoming speed is 9% by which bottle will move

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