Answer:
Given :
m 1 =0.5 kg
u 1 =0 m/s
m 2=0.005 kg
u 2=200 m/s
Let the velocity acquired by the block after the collision be V.
Using conservation of linear momentum before and after the collision : P i =P f
∴ m1u 1 +m 2u 2=(m 1+m 2 )V
OR
0.5×0+0.005×200=(0.5+0.005)V
⟹V≈1.9 m/s