Question

A ball of mass $100g$ is projected vertically upwards from the ground with a velocity of $49m/s$. At the same time another identical ball is dropped from a height of $98m$ to fall freely along the same path as followed by the first ball. After sometime the two balls collide and stick together. The velocity of the combined mass just after the collision is:

• A. $4.9m/supward$
• B. $4.9m/sdownward$
• C. $9.8m/supward$
• D. $49.8m/sdownward$

Answer 1

The correct option is

A

$4.9m/supward$

$4.9m/s$ Upward $\to$ Velocity of combined mass after collision

We know that, using relative motion

$S_{rel}=98m,u_{rel}=49-0m/s,a_{rel}=0m/s^2$

$\Rightarrow 98=(49-0)t+[1/2\times (0\left)t^2\right]$

$\Rightarrow t=\frac{98}{49}sec$

$\therefore t=2sec$

for the first ball which is thrown from ground, $v=u+at$

$v_1=u_1-9.8\left(2\right)$

$v_1=49-19.6$

$v_1=29.4m/sec$

for the second ball which is dropped from 98 m height simultaneously ,

$v_2=0+9.8\left(2\right)$

$v_2=19.6m/sec$

Now, using conservation of linear momentum ; $m_1{v}_1-m_2{v}_2=(m_1+m_2)V_{com}$

$\Rightarrow V_{com}=\frac{100\times {10}^{-3}\times (29.4-19.6)}{200\times {10}^{-3}}$

$\Rightarrow V_{com}=4.9m/secupward$