1. Initial momentum of the system

The first ball has mass $m_1=10\,\text{kg}$ and velocity $10\sqrt{3}\,\text{m/s}$ along the $+X$-axis.

So its momentum is

The second ball of mass $20\,\text{kg}$ is initially at rest, so its momentum is zero.

Hence total initial momentum:

2. After collision

It is given that the first ball comes to rest. So after collision, all momentum is carried by the two equal fragments of the second ball.

The second ball has mass $20\,\text{kg}$, so each fragment has mass

• Fragment 1 moves along the $Y$-axis with speed $10\,\text{m/s}$.
• Fragment 2 moves with speed $20\,\text{m/s}$ at angle $\theta$ with respect to the $X$-axis.

3. Momentum of each fragment

For fragment 1:

Since it moves along the $Y$-axis, its momentum vector is

(From the figure/configuration, this is along $+Y$.)

For fragment 2:

Its components are

4. Apply conservation of momentum

Total final momentum must equal total initial momentum:

So in components:

Along $X$-axis:

Along $Y$-axis:

Since initial $Y$-momentum is zero, the two fragments must cancel in $Y$. Thus the second fragment must have downward $Y$-component of magnitude $100$:

with sign determined by the figure (below the $X$-axis), so effectively $|\sin\theta|=\tfrac12$. Hence

for the acute angle with the $X$-axis.

5. Find $\theta$

From

we get

6. Nearest integer

7. Comparison with stored answer

Stored correct answer = $30$.

Our derived answer matches the stored answer.