Question

A ball of mass $100\text{g}$ is projected vertically upwards from the ground with a velocity of $49{\text{ms}}^{-1}$. At the same time, another identical ball is dropped from a height of $98\text{m}$ to fall freely along the same path as followed by the first ball. After sometimes the two balls collide and sticks together. The velocity of the combined mass just after the collision is-
[Take $g=9.8{\text{ms}}^{-2}$]

• A. $4.9{\text{ms}}^{-1}$
• B. $9.8{\text{ms}}^{-1}$
• C. $14.7{\text{ms}}^{-1}$
• D. $19.6{\text{ms}}^{-1}$

Answer 1

The correct option is A $4.9{\text{ms}}^{-1}$


Let balls collide at time $t.$

The height covered by the first ball in time $t$ is,

$x=ut+\frac{1}{2}g{t}^2$

$x=49t-\frac{1}{2}\left(9.8\right)t^2$

$x=49t-4.9t^2......\left(1\right)$

Displacement of second ball at time $t$ is,

$y=ut+\frac{1}{2}g{t}^2$

$y=0+\frac{1}{2}\times 9.8t^2[\because u=0]$

$y=4.9t^2............\left(2\right)$

From $\left(1\right)$ and $\left(2\right)$

$x+y=49t-4.9t^2+4.9t^2=98(\because x+y=98)$

$\therefore t=2\text{s}$

Velocity of ball $1$ at time of collision.

$v=u+at[\because a=-g]$

$\Rightarrow v_1=49-\left(9.8\right)\times 2$

$\Rightarrow v_1=29.4{\text{ms}}^{-1}$ (upwards)

Velocity of ball $2$ at time of collision is,

$v=u+at[\because a=g]$

$v=0+\left(9.8\right)\times 2=19.6{\text{ms}}^{-1}$(downwards)

Now, using conservation of linear momentum we get,

$[0.10\times 29.4]+[0.10\times (-19.6\left)\right]=0.20\times V$ (Taking upward as positive)

Where, $V$ is velocity after collision

$2.94-1.96=0.2\times V$

$\therefore V=4.9{\text{ms}}^{-1}$

<!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> Hence, $\left(\text{A}\right)$ is the correct answer.