1. Given data

• Mass of first ball: $m_1 = 10\,\text{kg}$
• Initial velocity of first ball: $u_1 = 10\sqrt{3}\,\text{m/s}$ along $+x$
• Mass of second ball: $m_2 = 20\,\text{kg}$, initially at rest
• After collision, first ball comes to rest.
• Second ball breaks into two equal pieces, so each piece has mass $\frac{20}{2} = 10\,\text{kg}.$
• One piece moves along $+y$ with speed $10\,\text{m/s}$.
• Other piece moves at $30^\circ$ with the $x$-axis with speed $x$.

We use conservation of linear momentum in both $x$ and $y$ directions.

2. Initial momentum

Initially, only the first ball is moving.

So total initial momentum is $\vec p_i = m_1 u_1 \hat i = 10 \cdot 10\sqrt{3}\,\hat i = 100\sqrt{3}\,\hat i.$

Thus,

• Initial $x$-momentum = $100\sqrt{3}$
• Initial $y$-momentum = $0$

3. Final momentum of pieces

After collision:

• First ball comes to rest, so its final momentum is $0$.

For the two fragments of the second ball:

Piece 1

Mass $=10\,\text{kg}$, velocity $=10\,\text{m/s}$ along $+y$.

Its momentum is $\vec p_{(1)} = 10 \cdot 10\,\hat j = 100\,\hat j.$

Piece 2

Mass $=10\,\text{kg}$, speed $=x$, direction at $30^\circ$ to the $x$-axis.

Since total initial $y$-momentum is zero, this piece must have a negative $y$-component to cancel the upward momentum of the first fragment.

So its momentum components are: $p_{(2)x} = 10x\cos 30^\circ,$ $p_{(2)y} = -10x\sin 30^\circ.$

4. Apply conservation of momentum in $y$-direction

Initial $y$-momentum is zero: $0 = 100 - 10x\sin 30^\circ.$

Using $\sin 30^\circ = \frac12$, $0 = 100 - 10x\cdot \frac12$ $0 = 100 - 5x$ $x = 20\,\text{m/s}.$

5. Check with $x$-direction

Initial $x$-momentum: $100\sqrt{3}$

Final $x$-momentum: $10x\cos 30^\circ = 10x\cdot \frac{\sqrt{3}}{2} = 5\sqrt{3}x$

Substitute $x=20$: $5\sqrt{3}(20)=100\sqrt{3}$

This matches the initial $x$-momentum, so the answer is consistent.

6. Final answer

To the nearest integer, $\boxed{20}$