a particle of mass m moving with a certain velocity collides elastically head on with a particle of mass 4m at rest. the percentage of K.E transferred is? (ans : 64%)

Dear Student,

Please find below the solution to the asked query:

By using conservation of momentum and conservation of energy, the equations for the velocities of the spheres after collision are,

$$\begin{gathered} V_1=\left(\frac{m_1-m_2}{m_1+m_2}\right)U_1+\left(\frac{2m_2}{m_1+m_2}\right)U_2 \\ and \\ {V}_2=\left(\frac{m_2-m_1}{m_1+m_2}\right)U_2+\left(\frac{2m_1}{m_1+m_2}\right)U_1 \end{gathered}$$

Where,
m₁ = mass of the first particle,
m₂ = mass of the second particle,
U₁ = speed of first particle before collision,
U₂ = speed of second particle before collision,
V₁ = speed of first particle after collision
V₂ = speed of second particle after collision

Therefore,

The kinetic energy transferred is equal to the kinetic energy of the second particle after collision. Therefore,

$$\begin{gathered} V_2=\left(\frac{m_2-m_1}{m_1+m_2}\right)U_2+\left(\frac{2m_1}{m_1+m_2}\right)U_1 \Rightarrow V_2=\left(\frac{4m-m}{m+4m}\right)\left(0\right)+\left(\frac{2m}{m+4m}\right)U \\ \Rightarrow V_2=\frac{2U}{5} \end{gathered}$$

So, the kinetic energy of the second particle after the collision is,

$$\begin{gathered} K.E = \frac{1}{2}\left(4m\right)V_2^2 = \frac{1}{2}\left(4m\right){\left(\frac{2U}{5}\right)}^2 \\ \Rightarrow K.E= \frac{16}{25}\left(\frac{1}{2}m{U}^2\right) \end{gathered}$$
Therefore, the percentage of the kinetic energy transferred is,

$$\begin{gathered} ∆K.E \% = \frac{K.E_2}{K.E_{i,tot}}\times 100 = \frac{\frac{16}{25}\left(\frac{1}{2}m{U}^2\right)}{\frac{1}{2}m{U}^2}\times 100 = \frac{16}{25}\times 100 \\ \Rightarrow ∆K.E \% = 64 \% \end{gathered}$$

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