Question

Two particles of masses $m_1$ and $m_2$ and velocities $u_1$ and $\alpha u_1(\alpha \ne 0)$ make an elastic bead on collision. If the initial kinetic energies of the two particles are equal and $m_1$ comes to rest after collision. Then

• A. $\frac{u_1}{u_2}=\sqrt{2}+1$
• B. $\frac{u_1}{u_2}=\sqrt{2}-1$
• C. $\frac{m_2}{m_1}=3+2\sqrt{2}$
• D. $\frac{m_2}{m_1}=3-2\sqrt{2}$

Answer 1

Answer: (B), (D)

$\frac{m_2}{m_1}=3-2\sqrt{2}$
$\frac{u_1}{u_2}=\sqrt{2}-1$

Two particle of mass $m_1$ and $m_2$ with velocity $u_1$ and $\alpha u_1$ respectively.

If the initial kinetic energy of two-particle is equal, and after the collision, $m_1$ comes at rest.

$\frac{1}{2}{m}_1{u}_1^2=\frac{1}{2}{m}_2(\alpha u_1{)}^2$

$\frac{m_1}{m_2}={\alpha }^2$. . . .(1)

By the conservation of linear momentum,

$m_1{u}_1+m_2\alpha u_1=m_2{v}_2$

$v_2=u_1\alpha (\alpha +1)$. . . .(2)

By the conservation of kinetic energy.

$\frac{1}{2}{m}_1{u}_1^2+\frac{1}{2}{m}_2(\alpha u_1{)}^2=\frac{1}{2}{m}_2{v}_2^2$

put equation (2) in the above equation, we get

$\frac{m_1}{m_2}={\alpha }^4-2{\alpha }^3={\alpha }^2$ (from equation 1)

${\alpha }^4-2{\alpha }^3-{\alpha }^2=0$

$\alpha =1+\sqrt{2}$

Put the value of $\alpha$ in equation (1), we get,

$\frac{m_1}{m_2}=3+2\sqrt{2}$

$\frac{m_2}{m_1}=3-2\sqrt{2}$. . . .(3)

The initial kinetic energy of particles is equal.

$\frac{1}{2}{m}_1{u}_1^2=\frac{1}{2}{m}_2{u}_2^2$

$(\frac{u_1}{u_2}{)}^2=\frac{m_2}{m_1}$

$(\frac{u_1}{u_2}{)}^2=3-2\sqrt{2}$

$\frac{u_1}{u_2}=\sqrt{2}-1$. . . .(4)

From equation (3) and (4) we conclude that the correct option is B and D.