Explanation
Mass of block 1 & 2, $${{m}_{1}}=10\,kg\,\,and\,{{m}_{2}}=5\,kg$$
Initial velocity 1 & 2, $${{v}_{1}}=30\,m{{s}^{-1}}\,\,and\,\,{{v}_{2}}=0$$
From conservation of momentum
Initial momentum = momentum of center of mass
$$ {{m}_{1}}{{u}_{1}}+{{m}_{2}}{{u}_{2}}=({{m}_{1}}+{{m}_{2}}){{V}_{cm}} $$
$$ \Rightarrow {{V}_{cm}}=\dfrac{{{m}_{1}}{{u}_{1}}+{{m}_{2}}{{u}_{2}}}{{{m}_{1}}+{{m}_{2}}}=\dfrac{10\times 30+5\times 0}{10+0}=20\,m{{s}^{-1}} $$
Hence, center of mass velocity is $$20\,m{{s}^{-1}}$$
Consider the point at which resultant of the forces will act, is 'x' m from the point A.
Thus the point is (1.5 - x) m far from the point B.
Thus, using the formula Torque = Distance x Force acting
$$20x = 30(1.5 - x)$$
or,$$ 2x = 4.5 - 3x$$
or,$$ 5x = 4.5$$
or, $$x = 0.9m or 90cm$$
Thus the resultant point will be 0.9 m or 90 cm far from point A
Two persons A and B of weight 80 kg and 50 kg respectively are standing at opposite ends of a boat of mass 70 kg and length 2m, at rest. When they interchange their positions then the displacement of the center of mass of the boat will be
Four uniform thin rods each of mass 1 kg and length 1 m are joined in the form of a square. If the square is rotated about axis AB, then it moment of inertia is
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