A stone is dropped from a height $$h.$$ It hits the ground with a certain momentum $$P.$$ If the same stone is dropped from a height $$100\% $$ more than the previous height, the momentum when it hits the ground will change by :
A.
$$68\% $$
B.
$$41\% $$
C.
$$200\% $$
D.
$$100\% $$
Answer :
$$41\% $$
Solution :
Momentum $$P = mv = m\sqrt {2gh} \,\,\left( {\because {v^2} = {u^2} + 2gh;\,{\text{Here}}\,u = 0} \right)$$
When stone hits the ground momentum
$$P = m\sqrt {2gh} $$
when same stone dropped from $$2h$$ ($$100\% $$ of initial) then momentum $$P' = m\sqrt {2g\left( {2h} \right)} = \sqrt 2 P$$
Which is changed by $$41\% $$ of initial.
Releted MCQ Question on Basic Physics >> Momentum
Releted Question 1
Two particles of masses $${m_1}$$ and $${m_2}$$ in projectile motion have velocities $${{\vec v}_1}$$ and $${{\vec v}_2}$$ respectively at time $$t = 0.$$ They collide at time $${t_0.}$$ Their velocities become $${{\vec v}_1}'$$ and $${{\vec v}_2}'$$ at time $$2{t_0}$$ while still moving in the air. The value of $$\left| {\left( {{m_1}{{\vec v}_1}' + {m_2}{{\vec v}_2}'} \right) - \left( {{m_1}{{\vec v}_1} + {m_2}{{\vec v}_2}} \right)} \right|$$ is
A.
zero
B.
$$\left( {{m_1} + {m_2}} \right)g{t_0}$$
C.
$$\frac{1}{2}\left( {{m_1} + {m_2}} \right)g{t_0}$$
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