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Situation 1: Ball B is at rest. 1.a. What will be the velocity of ball B after, if: the collision is perfectly elastic? You may use this formula: A(before)+B(before) = A(after)+B(after) (m)(v) +(m)(v) = (m)(v) +(m)(v)

1. b. What will be the velocity of ball B after, if the collision is perfectly inelastic? You may use this formula: A(before)+B(before) = A(after)+B(after) (m)(v) +(m)(v) = (mA + mB)(v)

Situation 2: Ball B travels twice the velocity of ball A. 2. a. What will be the velocity of ball B after, if: a. the collision is perfectly elastic? You may use this formula: A(before)+B(before) = A(after)+B(after) (m)(v) +(m)(v) = (m)(v) +(m)(v)

2.b. What will be the velocity of ball B after, if the collision is perfectly inelastic? You may use this formula. A(before)+B(before) = A(after)+B(after) (m)(v) +(m)(v) = (mA + mB) (v) ​

Situation 1 Ball B Is At Rest 1a What Will Be The Velocity Of Ball B After If The Collision Is Perfectly Elastic You May Use This Formula AbeforeBbefore AafterB class=

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Answer:-

[tex]\sf ☆m_A=4kg\\ ☆\sf m_B=4kg \\ ☆\sf u_1=2.5m/s \\ \sf ☆u_2=?[/tex]

According to law of conversion of linear momentum.

[tex]\boxed{\sf m_1u_1=m_2u_2}[/tex]

[tex]\sf\longmapsto 4\times 2.5=4\times u_2[/tex]

[tex]\sf\longmapsto 10=4u_2[/tex]

[tex]\sf\longmapsto u_2=\dfrac{10}{4}[/tex]

[tex]\sf\longmapsto u_2=2.5m/s[/tex]

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