A ${ }^{238}\mathrm{U}$ nucleus decays by emitting an alpha particle of speed $\mathrm{v}{\mathrm{ms}}^{-1}$. The recoil speed of the residual nucleus is (in  ${\mathrm{ms}}^{-1}$)

Initially ²³⁸U nucleus was at rest and after decay its part moves in opposite direction.

According to conservation of momentum
$4\mathrm{v}+234\mathrm{V}$= 238 × 0 $\Rightarrow \mathrm{V}=-\frac{4\mathrm{v}}{234}$