Question

At what value of $\eta$ will the velocity of the disc after the collision reverse its direction?

• A. $\eta <4$
• B. $\eta >4$
• C. $\eta =4$
• D. $\eta =0$

Answer 1

Answer: (B)

$\eta >4$

Let mass of the disc be one unit.

By conservation of linear momentum,

$v=v^{{}^{\prime }}+\eta v_o$

$v^{{}^{\prime }}$ is the velocity of disc after collision

$v_o=\frac{v-v^{{}^{\prime }}}{\eta }$

Conservation of angular momentum,

$v\frac{l}{2}=v^{{}^{\prime }}\frac{l}{2}+\frac{\eta l^2}{12}w$

$w$ is the angular velocity of rod.

$w=\frac{6(v-v^{{}^{\prime }})}{\eta l}$

Conservation of energy,

$\frac{1}{2}{v}^2=\frac{1}{2}{v}^{{}^{\prime }2}+\frac{1}{2}\eta v_o^2+\frac{1}{2}\frac{\eta l^2}{12}{w}^2$

Substituting $v_o$ and $w$ in energy equation

$(v^2-v^{{}^{\prime }2})=\eta (\frac{(v-v^{{}^{\prime }}{)}^2}{{\eta }^2}+3\frac{(v-v^{{}^{\prime }}{)}^2}{{\eta }^2})$

$(v^2-v^{{}^{\prime }2})=\frac{4(v-v^{{}^{\prime }}{)}^2}{\eta }$

Since $v\ne v^{{}^{\prime }}$

$v+v^{{}^{\prime }}=\frac{4(v-v^{{}^{\prime }})}{\eta }$

$v^{{}^{\prime }}=\frac{4-\eta }{4+\eta }v$

Thus we get

$v^{{}^{\prime }}=\frac{4-\eta }{4+\eta }v$

Therefore, $v^{{}^{\prime }}$ is negative when $\eta >4$
Answer: B