Center of Mass, Momentum & Collision

JEE Main level, single correct, medium

35 of 130

Question 35 of 130

The position of center of mass of three masses 22 kg, 33 kg and 1515 kg placed with respect to mid point (pp) of normal bisector, as shown in the figure is _______.

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Answer: (A) (34,1.25)\left(\dfrac{\sqrt{3}}{4}, 1.25\right)

Let the midpoint pp of the normal bisector be the origin (0,0)(0,0).

The triangle is isosceles with two sides of 10 m10\text{ m} and an included angle of 120∘120^\circ. The altitude from the 15 kg15\text{ kg} mass bisects the angle into two 60∘60^\circ angles.

The length of the altitude is 10cos⁡(60∘)=5 m10 \cos(60^\circ) = 5\text{ m}.

Since pp is the midpoint of this altitude, the 15 kg15\text{ kg} mass is at a distance of 2.5 m2.5\text{ m} above pp, and the base is 2.5 m2.5\text{ m} below pp.

The distance of the 2 kg2\text{ kg} and 3 kg3\text{ kg} masses from the altitude is 10sin⁡(60∘)=53 m10 \sin(60^\circ) = 5\sqrt{3}\text{ m}.

The coordinates of the masses with respect to pp are:
m1=15 kgm_1 = 15\text{ kg} at (0,2.5)(0, 2.5)
m2=2 kgm_2 = 2\text{ kg} at (−53,−2.5)(-5\sqrt{3}, -2.5)
m3=3 kgm_3 = 3\text{ kg} at (53,−2.5)(5\sqrt{3}, -2.5)

The x-coordinate of the center of mass is:
xcm=m1x1+m2x2+m3x3m1+m2+m3=15(0)+2(−53)+3(53)15+2+3=5320=34x_{cm} = \dfrac{m_1 x_1 + m_2 x_2 + m_3 x_3}{m_1 + m_2 + m_3} = \dfrac{15(0) + 2(-5\sqrt{3}) + 3(5\sqrt{3})}{15 + 2 + 3} = \dfrac{5\sqrt{3}}{20} = \dfrac{\sqrt{3}}{4}

The y-coordinate of the center of mass is:
ycm=m1y1+m2y2+m3y3m1+m2+m3=15(2.5)+2(−2.5)+3(−2.5)20=37.5−12.520=2520=1.25y_{cm} = \dfrac{m_1 y_1 + m_2 y_2 + m_3 y_3}{m_1 + m_2 + m_3} = \dfrac{15(2.5) + 2(-2.5) + 3(-2.5)}{20} = \dfrac{37.5 - 12.5}{20} = \dfrac{25}{20} = 1.25

The position of the center of mass is (34,1.25)\left(\dfrac{\sqrt{3}}{4}, 1.25\right).

Answer: (34,1.25)\left(\dfrac{\sqrt{3}}{4}, 1.25\right)

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