Question

One end of an ideal spring is fixed to a wall at origin $$O$$ and axis of spring is parallel to $$x-$$axis. $$A$$ block of mass $$m=1\ kg$$ is attached to free end of the spring and it is performing $$SHM$$. Equation of position of the block in coordinates system shown in figure is $$x=10+3\sin (10t)$$. Here $$t$$ is in second $$x$$ in $$cm$$.
Another blocks of mass $$M=3\ kg$$. moving towards the origin with velocity $$30\ cm/sec$$ collides with the block performing $$S.H.M$$ at $$t=0$$ and gets stuck to it.
Calculate:
New equation for position of the combined body,

Answer 1

Initially, $$\omega ^2 = \dfrac{k}{m}$$

At $$t=0$$, block of mass m is at mean position $$(x = 10m)$$ and moving towards positive x direction with velocity $$A\omega$$ or $$30 \ m/s$$

From the conservation of linear momentum,

$$(M+m)v = M(30)$$

Substituting the values, we have

$$v = $$ velocity of combined mass just after collision $$ = 15 \ cm/s \ or \ 0.15 m$$

From the conservation of mechanical energy,

$$\dfrac{1}{2}(M+m)v^2 = \dfrac{1}{2}k(A)^2$$

Here $$A$$ is the new amplitude os oscillation

$$A = v\sqrt{\dfrac{M+m}{k}} = 0.15\sqrt{\dfrac{4}{100}} = 0.03 \ m $$

$$\therefore A = 30\ cm$$

Now, new angular frquency, $$\omega = \sqrt{\dfrac{k}{M+m}} = 5\ rad/s$$

$$\therefore Equation: x = 10 - 3sin(5t)$$

[$$NOTE:$$ there is a minus sign because after collition direction will be changed]