Question

One end of an ideal spring is fixed to a wall at origin $O$ and the axis of spring is parallel to x-axis. A block of mass $m=1$ kg is attached to free end of the spring and its performing SHM. Equation of position of block in coordinate system shown is $x=10+sin10t$, $t$ is in second and $x$ in cm. Another block of mass $M=3$ kg, moving towards the origin with velocity $30cm/s$ collides with the block performing SHM at $t=0$ and gets stuck to it, calculate:(i) new amplitude of oscillations (ii) new equation for position of the combined body. (iii) loss of energy during collision. Neglect friction

• A. $4cm,x=10-4sin5t,\Delta E=0.06J$
• B. $5cm,x=10-3sin5t,\Delta E=0.135J$
• C. $3cm,x=5-3sin5t,\Delta E=0.135J$
• D. $3cm,x=10-3sin5t,\Delta E=0.270J$

Answer 1

The correct option is A $4cm,x=10-4sin5t,\Delta E=0.06J$

In this question $x=10+\sin 10t$

Here $\omega$ = 10 = =

so $k=100N/m$

now differentiate this equation to get velocity of the block at the time t= 0

so $v=10\cos 10t$

so at t = 0

v = 10 $\cos 0$

v = 10 cm/s

moving in the direction opposite to that of 3 kg block

so when they collide momentum will be conserved

so initial momentum = final momentum

so

so $v^{{}^{\prime }}=-20$

Now $w^{{}^{\prime }}=\sqrt{k/m}$

$w^{{}^{\prime }}=\sqrt{100/4}$

$w^{{}^{\prime }}=5$

$v^{{}^{\prime }}=A^{{}^{\prime }}{w}^{{}^{\prime }}$

so = 4 cm

so equation of motion is $x=A-a\sin wt=10-4\sin 5t$

loss of energy during the collision = final kinetic energy - intial kinetic energy

=$1/2\times (1+3)\times \left(0.2\right)\times \left(0.2\right)-1/2\times \left(1\right)\times \left(0.1\right)\times \left(0.1\right)-1/2\times \left(3\right)\left(0.3\right)\left(0.3\right)$

$=0.08-0.005-0.135$

$=0.06J$