Center of Mass, Momentum & Collision

JEE Main level, numerical answer, medium

113 of 130

Question 113 of 130

A body AA of mass m=0.1kgm=0.1kg has an initial velocity of 3i^ms−13\hat{i}m{s}^{-1}. It collides elastically with another body BB of the same mass which has an initial velocity of 5j^ms−15\hat{j}m{s}^{-1}. After the collision, AA moves with a velocity v⃗=4(i^+j^)ms−1\vec{v}=4\left(\hat{i}+\hat{j}\right)m{s}^{-1}. The energy of BB after the collision is written as x10J\dfrac{x}{10}J. The value of xx is

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Answer: 1

Using the principle of conservation of linear momentum
0.1×3i^+0.1×5j^=0.1×4(i^+j^)+0.1×v⃗0.1\times 3\hat{i}+0.1\times 5\hat{j}=0.1\times 4\left(\hat{i}+\hat{j}\right)+0.1\times \vec{v}
v⃗=(−i^+j^)ms−1\vec{v}=\left(-\hat{i}+\hat{j}\right)m{s}^{-1}
v=∣v⃗∣=2ms−1v=\left|\vec{v}\right|=\sqrt{2}m{s}^{-1}
KB=12×0.1×v2=110J{K}_{B}=\dfrac{1}{2}\times 0.1\times {v}^{2}=\dfrac{1}{10}J
x=1x=1

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