Answer
$21\space g,\space x(+)\space direction$
Work Step by Step
Please see the attached image below.
Here we use the principle of conservation of momentum.
$ m_{1}u_{1}+m_{2}u_{2}=m_{1}V_{1}+m_{2}V_{2}$
Let's apply this principle in both x, and y direction as follows.
$\rightarrow m_{1}u_{1}=m_{2}u+m_{3}u_{3}$
$0=m_{2}\times u+14\space g\times 48\space m/s$
$-672\times10^{-8}kgm/s=m_{2}u-(1)$
$\uparrow 0= m_{2}V+14\space g\times (0)=\gt V=0$
Given that, the speed of second mass =32 m/s
So, we can write $=\gt 32\space m/s=\sqrt {u^{2}+V^{2}}=u$
Direction of the second particle = x(+) direction
$(1)=\gt \frac{-672\times10^{-3}kgm/s}{32\space m/s}=m_{2}$
$21 g =m_{2}$
