A ball of mass 10 kg moving with a velocity \(10 \sqrt{3} \mathrm{~ms}^{-1}\) along \(\mathrm{X}-\) axis, hits another ball of mass 20 kg which is at rest. After collision, the first ball comes to rest and the second one disintegrates into two equal pieces. One of the pieces starts moving along Y - axis at a speed of \(10 \mathrm{~m} / \mathrm{s}\). The second piece starts moving at a speed of \(20 \mathrm{~m} / \mathrm{s}\) at an angle \(\theta\) (degree) with respect to the \(X\)-axis. The configuration of pieces after collision is shown in the figure. The value of \(\theta\) to the nearest integer is

JEE Main · 2021 · 16 Mar 2021 · Shift 1 · Conservation of Momentum

A ball of mass 10 kg moving with a velocity

103 ms^{- 1}

along

X -

axis, hits another ball of mass 20 kg which is at rest. After collision, the first ball comes to rest and the second one disintegrates into two equal pieces. One of the pieces starts moving along Y - axis at a speed of

10 m / s

. The second piece starts moving at a speed of

20 m / s

at an angle

θ

(degree) with respect to the

X

-axis.
The configuration of pieces after collision is shown in the figure. The value of

θ

to the nearest integer is

Answer: 30

Solution

The first ball comes to rest, so the whole of its momentum passes to the two 10 kg fragments.
Along the

y

-axis the incoming momentum is zero, and the piece travelling up the

y

-axis carries

10 \times 10 = 100 kg m s^{- 1}

. The second piece, directed below the

x

-axis, must cancel it:

10 \times 20 sin θ = 100 \Rightarrow sin θ = \frac{1}{2}

The

x

-axis balance confirms this angle, since

10 \times 20 cos 3 0^{\circ} = 1003 kg m s^{- 1}

equals the incoming momentum

10 \times 103

.
The second piece moves at

θ = 30

degrees below the

x

-axis.

JEE Main · 2021 · 16 Mar 2021 · Shift 1 · Integer answer

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JEE Main · 2021 · 16 Mar 2021 · Shift 1 · Conservation of Momentum

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