Question

Two balls of equal masses are thrown upwards along the same vertical line at an interval of $2$ seconds with the same initial velocity of $39.2m{s}^{-1}$. The total time of flight of each ball, if they collide at a certain height, and the collision is perfectly inelastic, will be

• A. $5s$ and $3s$
• B. $10s$ and $6s$
• C. $5\sqrt{15}s$ and $3\sqrt{15}s$
• D. $(5+\sqrt{15})s$ and $(3+\sqrt{15})s$

Answer 1

The correct option is D $(5+\sqrt{15})s$ and $(3+\sqrt{15})s$

When the particles collide the height of both particles will be same,
For first particle time will be $t$ and for second particle time will be $t-2$
$h=ut-\frac{1}{2}g{t}^2=u(t-2)-\frac{1}{2}g{(t-2)}^2$

$ut-\frac{1}{2}g{t}^2=ut-2u-\frac{1}{2}g(t^2-4t+4)$

$0=-2u+2gt-2g$

$t=\frac{u}{g}+1$

$t=5s$

By solving the quadratic equation in $t$ we get the time of first collision to be $5$s.

$h=39.2\times 5-\frac{1}{2}g\times 5^2=20g-12.5g=7.5gmeter$

Now by momentum conservation, velocity of particles after collision

$m(u-gt)+m(u-g(t-2\left)\right)=m{v}^{\prime }$

$mu-5mg+mu-3mg=m{v}^{\prime }$

$v^{\prime }=2u-8g=0m/s$

$a=-g$

Now for collision with ground let time taken be $t_2$

$h=\frac{1}{2}g{t}_2^2\Rightarrow t=\sqrt{\frac{2h}{g}}=\sqrt{\frac{2\times 7.5g}{g}}=\sqrt{15}$

Thus, times of total time of flights will be $5+\sqrt{15}$s and $3+\sqrt{15}$s