Question

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

• A. $4.9$ m/s upward
• B. $4.9$ m/s downward
• C. $9.8$ m/s upward
• D. $9.8$ m/s downward

Answer 1

The correct option is A $4.9$ m/s upward

time after which that meet$,$

$t=\frac{98}{49}s=2s$

velocity of first ball and second ball at time of collision

$v_1=49-2gand{v}_2=2g$

$\begin{array}{c}2m\left(v\right)=m_1{v}_1+m_2{v}_2\\ \Rightarrow 2m\left(v\right)=m\left(49-2g\right)+m\left(-2g\right)\\ \Rightarrow 2v=49-2g-2g\\ 2v=49-4g\\ 2v=49-4\times 9.8\\ v=4.9m/supwards.\end{array}$

Hence,

option $\left(A\right)$ is correct answer.