Question

A block of mass $M=40\text{kg}$ is released on a smooth incline from the point shown in the figure. After travelling through a length of $30\text{cm}$, it strikes an ideal spring of force constant $k=1000\text{N/m}$. It compresses the spring and then gets pushed back. How much time after its release will the block be back again to the point of release?
[Take $g=10{\text{m/s}}^2$]

• A. $0.70\text{s}$
• B. $0.84\text{s}$
• C. $0.46\text{s}$
• D. $1.54\text{s}$

Answer 1

The correct option is D $1.54\text{s}$

Time to travel from A to B $\left(t_1\right)$ is given by -
$x=ut+\frac{1}{2}a{t}^2$
$[\because a=g\sin \theta =5{\text{m/s}}^2]$
$0.3=\frac{1}{2}\times 5\times t_1^2\Rightarrow t_1=0.35\text{s}$
Speed of the block when it hits the spring is given by -
$V^2=0^2+2ax=2\times 5\times 0.3$
$V=\sqrt{3}\text{m/s}$
The motion of the block in contact with the spring can be regarded as part of an SHM. In equilibrium, compression in the spring is given by
$k{X}_0=Mg\sin \theta \Rightarrow X_0=\frac{40\times 10\times \frac{1}{2}}{1000}=0.2\text{m}$
Let the block compress the spring by a maximum distance $X_1$ (when it comes to rest).
Energy conservation gives -
$\frac{1}{2}k{X}_1^2=\frac{1}{2}M{V}^2+Mg{X}_1\sin \theta$
$\frac{1}{2}\times 1000X_1^2=\frac{1}{2}\times 40\times (\sqrt{3}{)}^2+(40\times 10\times \frac{1}{2})X_1$
$5X_1^2-2X_1-0.6=0$
Solving, $X_1=0.6\text{m}$
Hence, amplitude of SHM is $A=X_1-X_0=0.4\text{m}$


So, $X_o=A/2$
The motion from $B$ to $C$ and back to $B$ can be regarded as motion of a particle performing SHM from $x=+\frac{A}{2}$ to negative extreme and back to $x=+\frac{A}{2}$
Time for this motion can be obtained from the phase diagram given below.


Particle on phase diagram moves from $P_1$ to $C$ to $P_2.$
Time required for completing this two third circle $=\frac{2T}{3}$
Hence, total time for the part $B\to C\to B$
$t_2=\frac{2T}{3}=\frac{2}{3}\times 2\pi \sqrt{\frac{m}{k}}$$=\frac{4\pi \sqrt{40}}{3\sqrt{1000}}=0.84\text{s}$
$\therefore$ Total time required for the block to come back to A is
$t=2t_1+t_2$ [by symmetry, since block leaves the spring with the same velocity it hits it with]
$=2\times 0.35+0.84=1.54\text{s}$