Center of Mass, Momentum & Collision

JEE Main level, single correct, medium

66 of 130

Question 66 of 130

Two identical bodies AA and BB of equal masses have initial velocities v1⃗=4i^\vec{v_1} = 4\hat{i} m/s and v2⃗=4j^\vec{v_2} = 4\hat{j} m/s respectively. The body AA has acceleration a1⃗=6i^+6j^\vec{a_1} = 6\hat{i} + 6\hat{j} m/s2^2 while the acceleration of the other body BB is zero. The centre of mass of the two bodies moves in __________ path.

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Answer: (C) straight line

Velocity of the centre of mass is given by:

v⃗cm=m1v⃗1+m2v⃗2m1+m2\vec{v}_{cm} = \dfrac{m_1\vec{v}_1 + m_2\vec{v}_2}{m_1 + m_2}

Since m1=m2=mm_1 = m_2 = m, we get:

v⃗cm=m(4i^)+m(4j^)2m=2i^+2j^\vec{v}_{cm} = \dfrac{m(4\hat{i}) + m(4\hat{j})}{2m} = 2\hat{i} + 2\hat{j} m/s

Acceleration of the centre of mass is given by:

a⃗cm=m1a⃗1+m2a⃗2m1+m2\vec{a}_{cm} = \dfrac{m_1\vec{a}_1 + m_2\vec{a}_2}{m_1 + m_2}

a⃗cm=m(6i^+6j^)+02m=3i^+3j^\vec{a}_{cm} = \dfrac{m(6\hat{i} + 6\hat{j}) + 0}{2m} = 3\hat{i} + 3\hat{j} m/s2^2

Here, v⃗cm=2(i^+j^)\vec{v}_{cm} = 2(\hat{i} + \hat{j}) and a⃗cm=3(i^+j^)\vec{a}_{cm} = 3(\hat{i} + \hat{j}).

Since the initial velocity and acceleration of the centre of mass are parallel to each other, the centre of mass moves in a straight line path.

Answer: straight line

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