Ans.

A particle of mass \( m \) moving with velocity \( U \) collides with a particle at rest.
Let us consider, \( m_2 \) be the velocity of the target particle.

After collision, the target particle moves forward with velocity \( v_2=\frac{u}{3} \) and the incident particles moves backward with velocity \( v_1=-\frac{2u}{3} \)

According to the conservation of linear momentum,

or, \( mu=\frac{2mu}{3}-\frac{m_2u}{3} \)

or, \( mu\left(1+\frac{2}{3}\right)=\frac{m_2u}{3} \)

or, \( \frac{5mu}{3}=\frac{m_2u}{3} \)

or, \( m_2=5m \)

So the mass of the target particle is 5m.