A body A , of mass $\mathrm{m}=0.1 \mathrm{~kg}$ has an initial velocity of $3 \hat{\mathrm{i}} \mathrm{ms}^{-1}$. It collides elastically with another body, $B$ of the same mass which has an initial velocity of $5 \hat{j} \mathrm{ms}^{-1}$. After collision, A moves with a velocity $\vec{v}=4(\hat{i}+\hat{j})$. The energy of $B$ after collision is written as $\frac{x}{10} \mathrm{~J}$. The value of $x$ is \_\_\_\_ .

JEE Main · 2020 · 08 Jan 2020 · Shift 1 · Conservation of Momentum

A body A , of mass

m = 0.1 kg

has an initial velocity of

3 \hat{i} ms^{- 1}

. It collides elastically with another body,

B

of the same mass which has an initial velocity of

5 \hat{j} ms^{- 1}

. After collision, A moves with a velocity

v = 4 (\hat{i} + \hat{j})

. The energy of

B

after collision is written as

\frac{x}{10} J

. The value of

x

is ____ .

Answer: 1

Solution

An elastic collision conserves kinetic energy. Initially

K E_{i} = \frac{1}{2} m (3^{2}) + \frac{1}{2} m (5^{2}) = \frac{1}{2} m (9 + 25) = 17 m

. Afterwards particle

A

carries

K E_{A} = \frac{1}{2} m ∣4 (\hat{i} + \hat{j}) ∣^{2} = \frac{1}{2} m (16 \cdot 2) = 16 m

. Conservation gives

17 m = 16 m + K E_{B}

, so

K E_{B} = m = 0.1

= \frac{1}{10}

J, and therefore

x = 1

JEE Main · 2020 · 08 Jan 2020 · Shift 1 · Integer answer

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