Question

Two bodies A and B of masses m and 2m respectively are placed on a smooth floor.They are connected by a spring .A third body C of mass m moves with velocity $v_0$ along the line joining A and B and collides elastically with A as shown in fig.At a certain instant of time $t_0$ after collision , it is found that the instantaneous velocities of A and B are the same.Further at this instant the compression of the spring is found to be $x_0$ .Determine -(i) the common velocity of A and B at time $t_0$ and (ii) the spring constant.

Answer 1

By momentum conservation $A$ and $C$ will get rest and $A$ acquire velocity $v_0$.

$\left(i\right)$ When $A$ and $B$ both have same velocity then by linear momentum conservation, $m{v}_0=mv+3mv$

$v=\frac{v_0}{4}$

$\left(ii\right)$ Potential energy stored in spring= Kinetic energy changes

$\frac{1}{2}k{x}_0^2=\frac{1}{2}m{v}_0^2-\frac{1}{2}m(\frac{v_0}{4}{)}^2+\frac{1}{2}3m(\frac{v_0}{4}{)}^2$

$k{x}_0^2=m{v}_0^2(1-\frac{1}{16}+\frac{3}{16})$

$=m{v}_0^2\left(\frac{12}{16}\right)$

$k{x}^2=\frac{3}{4}m{v}_0^2$

$k=\sqrt{\frac{3}{4}m}\frac{v^0}{x^0}$