Question

A 5 kg ball moving at 5 ms−1strikes a 2 kg ball at rest. If the collision is elastic, what is the speed of each ball after the collision?

Answer 1

$\huge\underline\blue{\sf Solution:}$

$\large\underline\pink{\sf Given: }$

• Mass of ball $\sf{(m_1)=5kg}$

• Velocity of ball of mass 5kg $\sf{(u_1)=5m/s}$

• Mass of ball $\sf{(m_2)=2kg}$

• Velocity of ball of mass 2kg $\sf{(u_2)=0m/s}$

$\large\underline\pink{\sf To\:Find: }$

• Speed of each balls after Collision $\sf{(v_1)\:and\:(v_2)}$= ?

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According to law of conservation of momentum for elastic Collision:

$\large{\boxed{\sf m_1v_1+m_2v_2=m_1u_1+m_2u_2}}$

$\large\implies{\sf 5v_1+2v_2=5×5+2×0}$

$\large\implies{\sf 5v_1+2v_2=25→(1)}$

Applying equation of coefficient of restitution , we have :

$\large{\boxed{\sf v_2-v_1=e(u_1-u_2) }}$

$\large\implies{\sf v_2-v_1=e(5-0)}$

$\large\implies{\sf v_2-v_1=5e}$

For elastic Collision e=1

$\large\implies{\sf v_2-v_1=5→(2)}$

From Equation 3 and Equation 1

$\large\implies{\sf v_2=5+v_1}$

$\large\implies{\sf 5v_1+2(5+v_1)=25}$

$\large\implies{\sf 5v_1+10+2v_1=25}$

$\large\implies{\sf 10+7v_1=25}$

$\large\implies{\sf 7v_1=15 }$

$\large\implies{\sf v_1=2.15m/s }$

Now ,

Putting value of $\sf{v_1}$ in Equation 2

$\large\implies{\sf v_2-2.15=5}$

$\large\implies{\sf v_2=7.5m/s}$

Hence ,

Velocity of ball of mass 5kg $\sf{(v_1)}$ is 2.15m/s

and

Velocity of ball of mass 2kg $\sf{(v_2)}$ is 7.5m/s