Question
A particle of mass $$1\,kg$$ is thrown vertically upwards with speed $$100\,m/s.$$ After $$5\,s,$$ it explodes into two parts. One part of mass $$400\,g$$ comes back with speed $$25\,m/s,$$ what is the speed of other part just after explosion?
A.
$$100\,m/s\,{\text{upwards}}$$
B.
$$600\,m/s\,{\text{upwards}}$$
C.
$$100\,m/s\,{\text{downwards}}$$
D.
$$300\,m/s\,{\text{upwards}}$$
Answer :
$$100\,m/s\,{\text{upwards}}$$
Solution :
According to 1st equation of motion, velocity of particle after $$5\,s$$
$$\eqalign{
& v = u - gt \cr
& v = 100 - 10 \times 5 \cr
& = 100 - 50 = 50\,m/s\,\,\left( {{\text{upwards}}} \right) \cr} $$
Applying conservation of linear momentum gives
$$Mv = {m_1}{v_1} + {m_2}{v_2}\,.......\left( {{\text{i}}} \right)$$
Taking upward direction positive, the velocity $${v_1}$$ will be negative.
$$\eqalign{
& \therefore {v_1} = - 25\,m/s, \cr
& v = 50\,m/s \cr} $$
$$\eqalign{
& {\text{Also,}}\,\,M = 1\,kg,\,{m_1} = 400\,g = 0.4\,kg \cr
& {\text{and}}\,\,\,{m_2} = \left( {M - {m_1}} \right) = 1 - 0.4 = 0.6\,kg \cr} $$
Thus, Eq. (i) becomes,
$$\eqalign{
& 1 \times 50 = 0.4 \times \left( { - 25} \right) + 0.6{v_2} \cr
& {\text{or}}\,\,50 = - 10 + 0.6{v_2} \cr} $$
$$\eqalign{
& {\text{or}}\,\,0.6{v_2} = 60 \cr
& {\text{or}}\,\,{v_2} = \frac{{60}}{{0.6}} \cr
& = 100\,m/s \cr} $$
As $${v_2}$$ is positive, therefore the other part will move upwards with a velocity of $$100\,m/s.$$