The mass of the ball is\(m\).

For the first ball, let the initial speed be\(u\)and the final speed be\(u'\).

The initial speed of the second ball is\(v = 0{\rm{ m/s}}\).

The rebound speed of the first ball is\(u' = - 0.450u\).

The relation for relative velocities in a perfectly elastic collision is

\(\begin{array}{c}u - v = - \left( {u' - v'} \right)\\u - 0 = - \left( {u' - v'} \right)\\v' = \left( {u + u'} \right).\end{array}\) … (i)

Here, \(v'\) is the rebound speed of the second ball.

As no external force acts on the system of balls, the momentum remains conserved.

The relation obtained from the momentum conservation equation is

\(\begin{array}{c}P = P'\\mu = mu' + m'v'.\end{array}\)

Here, m is the mass of the first ball, and\[m'\)is the mass of the second ball.

Plugging the values in the above equation, you get

\(\begin{array}{c}mu = m\left( { - 0.450u} \right) + m'\left( {u + u'} \right)\\mu = m\left( { - 0.450u} \right) + m'\left( {u - 0.450u} \right)\\mu = m\left( { - 0.450u} \right) + m'\left( {0.550u} \right)\\m' = 2.64m.\end{array}\)

Thus, \(m' = 2.64m\) is the required mass of the second ball.