Two blocks of masses $$10kg$$ and $$4kg$$ are connected by a spring of negligible mass and placed on a frictionless horizontal surface. An impulse gives a velocity of $$14 m/s$$ to the heavier block in the direction of the lighter block. The velocity of the centre of mass is
A.
$$30 m/s$$
B.
$$20 m/s$$
C.
$$10 m/s$$
D.
$$5 m/s$$
Answer :
$$10 m/s$$
Solution :
Just after collision
$${v_c} = \frac{{{m_1}{v_1} + {m_2}{v_2}}}{{{m_1} + {m_2}}} = \frac{{10 \times 14 + 4 \times 0}}{{10 + 4}} = 10\,m/s;$$ Note : Spring force is an internal force, it cannot change the linear momentum of the (two mass + spring) system. Therefore $${v_c}$$ remains the same.
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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