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A billiard ball (ball #1) moving at 5.00 m/s strikes a stationary ball (ball #2) of the same mass. after the collision, ball #1 moves at a speed of 4.35 m/s. find the speed of ball #2 after the collision.

Respuesta :

if it is completely elastic, you can calculate the velocity of the second ball from the kinetic energy 
v1 = velocity of #1 
v1' = velocity of #1 after collision 
v2' = velocity of #2 after collision. 

kinetic energy: v1^2 = v1' ^2 + v2' ^2 (1/2 and m cancel out) 
5^2 = 4.35^2 + v2' ^2 
v2 = 2.46 m/s <--- ANSWER
Cricetus

After collision, "2.47 m/s" would be the speed of the 2nd ball.

According to the question,

Before collision,

  • Speed of 1st ball = 5.00 m/s

After collision,

  • Speed of 1st ball = 4.35 m/s

By using the energy conservation, we get

→ [tex]E_f = \frac{1}{2} mv_1^2+\frac{1}{2}mv_2^2[/tex]

and,

→ [tex]\frac{1}{2} mv^2= \frac{1}{2} mv_1^2+\frac{1}{2} mv_2^2[/tex]

or,

→      [tex]v^2=v_1^2+v_2^2[/tex]

        [tex]v_2 = \sqrt{(v^2-v_1^2)}[/tex]

By substituting the values, we get

            [tex]= \sqrt{(5^2-4.25^2)}[/tex]

            [tex]= 2.47 \ m/s[/tex]

Thus the above answer is right.                    

Learn more about collision here:

https://brainly.com/question/14377491

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