Discussion Forum

1)

 A block of mass ($10 \alpha$) g, where $\alpha$ is a constant is moving with velocity 3m/s to the right collides inelastically with the block on the right with mass 10 g and sticks to it. The right block is connected to three springs as shown in the figure. The spring constant of each spring is 2 N/m. if the amplitude  of the resulting simple harmonic motion is $\frac{1}{2\sqrt{2}}$ m, then the value of $\alpha$ is 

Answer:

Option A

Explanation:

 Equivalent spring constant  of  system of springs =$\frac{3k}{2}=3 N/m$

 Momentum is conserved in collision. Let blocks moves with velocity v just after collision  then

 $m_{1}v_{1}=(m_{1}+m_{2})v$

 $ 10 \alpha  \times 10^{-3} \times 3=(10 \alpha +10)\times 10^{-3} \times v$

$\Rightarrow$    $v= \frac{30 \alpha}{10(\alpha+1)}= \frac{3 \alpha}{\alpha+1} ms^{-1}$

 Displacement of blocks is  $\frac{1}{2\sqrt{2}}$m.

 so , energy conservation  gives, 

$\frac{1}{2}(m_{1}+m_{2})v^{2}=\frac{1}{2}k_{eq} x^{2}$ [ friction is absent]

 $\Rightarrow$   $\frac{(10\alpha+10)\times 10^{-3}\times 9\alpha^{2}}{(\alpha+1)^{2}}=3\times \frac{1}{8}$

$\Rightarrow$  $\frac{ 9\alpha^{2}}{100(\alpha+1)^{}}=\frac{3}{8}$

 $24 \alpha^{2}=100 \alpha +100 \Rightarrow \alpha =5$