Question
Three different objects of masses $${m_1},{m_2}$$ and $${m_3}$$ are allowed to fall from rest and from the same point $$O$$ along three different frictionless paths. The speeds of the three objects on reaching the ground will be in the ratio of
A.
$${m_1}:{m_2}:{m_3}$$
B.
$${m_1}:2{m_2}:3{m_3}$$
C.
$$1:1:1$$
D.
$$\frac{1}{{{m_1}}}:\frac{1}{{{m_2}}}:\frac{1}{{{m_3}}}$$
Answer :
$$1:1:1$$
Solution :
When an object falls freely under gravity, then its speed depends only on its height of fall and is independent of the mass of the object. As all objects are falling through the same height, therefore their speeds on reaching the ground will be in the ratio of $$1:1:1.$$
Alternative
The vertical displacement for all the three is same and paths are frictionless. So, by conservation of mechanical energy,
$$\eqalign{ & \frac{1}{2}m{v^2} = mgl \Rightarrow v = \sqrt {2gl} \cr & {\text{So,}}\,{v_1}:{v_2}:{v_3} = 1:1:1 \cr} $$
When an object falls freely under gravity, then its speed depends only on its height of fall and is independent of the mass of the object. As all objects are falling through the same height, therefore their speeds on reaching the ground will be in the ratio of $$1:1:1.$$
Alternative
The vertical displacement for all the three is same and paths are frictionless. So, by conservation of mechanical energy,
$$\eqalign{ & \frac{1}{2}m{v^2} = mgl \Rightarrow v = \sqrt {2gl} \cr & {\text{So,}}\,{v_1}:{v_2}:{v_3} = 1:1:1 \cr} $$
