Question

Two identical blocks P and Q have mass ‘m’ each. They are attached to two identical springs initially unstretched. Now the left spring (along with P) is compressed by $\frac{A}{2}$ and the right spring (along with Q) is compressed by A. Both the blocks are released simultaneously. They collide perfectly inelastically. Initial time period of both the blocks was T. [consider k as spring constant of each spring]

The time period and amplitude of oscillation of the combined mass is respectively,

• A. $\frac{T}{\sqrt{2}}$ and $\frac{A}{4}$
• B. $\frac{T}{2}$ and $\frac{A}{2}$
• C. $T$ and $\frac{A}{4}$
• D. $\frac{T}{2}$ and $\frac{A}{4}$

Answer 1

The correct option is C

$T$ and $\frac{A}{4}$

Time Period:
They will collide at their mean positions because time period of both are same and that is $2\pi \sqrt{\frac{m}{k}}$. After collision combined mass is 2m and $k_{eff}=2k$. Hence, time period remains unchanged.
Amplitude:
Applying conservation of linear momentum, velocity of combined mass just after collision is $v=\frac{A\omega }{4}$

Since this is the velocity at mean position,
Hence, $v={\omega }^{{}^{\prime }}{A}^{{}^{\prime }}$
or $\frac{\omega A}{4}={\omega }^{{}^{\prime }}{A}^{{}^{\prime }}$
or $A^{{}^{\prime }}=\frac{A}{4}as{\omega }^{{}^{\prime }}=\omega =\sqrt{\frac{k}{m}}=\sqrt{\frac{2k}{2m}}$