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13.Oscillations
normal
A spring-block system is resting on a frictionless floor as shown in the figure. The spring constant is $2.0 \mathrm{~N} \mathrm{~m}^{-1}$ and the mass of the block is $2.0 \mathrm{~kg}$. Ignore the mass of the spring. Initially the spring is in an unstretched condition. Another block of mass $1.0 \mathrm{~kg}$ moving with a speed of $2.0 \mathrm{~m} \mathrm{~s}^{-1}$ collides elastically with the first block. The collision is such that the $2.0 \mathrm{~kg}$ block does not hit the wall. The distance, in metres, between the two blocks when the spring returns to its unstretched position for the first time after the collision is. . . .
(image)
A$2.07$
B$2.08$
C$2.09$
D$2.10$
(IIT-2018)
Solution
After collision$1 v_1+2 v_2=1 \times 2$
$v_1+2 v_1=2$ $. . . . (i)$
(image)
$e=\frac{v_2-v_1}{2}=1$
$\quad v_2-v_1=2$ $. . . . (ii)$
On solving $\mathrm{v}_2=\frac{4}{3} \mathrm{~ms}^{-1}$ and $\mathrm{v}_1=-\frac{2}{3} \mathrm{~m} \mathrm{~s}^{-1}$
Time period $\mathrm{T}=2 \pi \sqrt{\frac{\mathrm{m}}{\mathrm{k}}}=2 \pi \sqrt{\frac{2}{2}}=2 \pi$
Displacement of $1^{\text {st }}$ block in time $\Delta \mathrm{t}=\frac{\mathrm{T}}{2}$
$\Delta \mathrm{s}=\frac{2}{3} \times \pi=\frac{6.28}{3}=2.09$
$\Delta \mathrm{s}=2.09 \mathrm{~m}$
Standard 11
Physics
