Question

A ball of mass $100 \,g$ is projected vertically upwards 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 sometime 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

Answer: (A)

Let balls collide at time $t$,
they are at same height $h$.
Height of first ball after $t$ seconds
$=49 t-0.5\left(9.8 t^{2}\right)$
$=4.9 t(10-t)$
Height of second ball after $t$ seconds
$=98$ downward distance moved by it in $t$ seconds
$=98-0.5(9.8) t^{2}=4.9\left(20-t^{2}\right) $
$\Rightarrow $ $4.9(10-t)=4.9\left(20-t^{2}\right) $
$\Rightarrow 10 t-t^{2}=20-t^{2} $
$\Rightarrow t=2 \,s$
The balls thus collide after $2\, s$ the start of their motion.
Their velocities at this instance are Ball
$1 : v_{1}=(49-98 \times 2)=29.4\, m / s$
upward Ball $2: v_{2}=(0+9.8 \times 2)=19.6 \,m / s$
downwards From conservation of momentum
$200 \,v=100 \times 29.4-100 \times 19.6$
$ \Rightarrow $ $v=4.9\, m / s$