A 500kg boat has an initial speed of 10ms^–1 as it passes under a bridge. At that instant a 50 kg man jumps straight down into the boat from the bridge. The speed of the boat after the man and boat attain a common speed is

We know from the law of conservation of momentum
${\mathrm{m}}_1{\mathrm{v}}_1={\mathrm{m}}_2{\mathrm{v}}_2$
Here  ${\mathrm{m}}_1$=500 Kg.
${\mathrm{v}}_1$​=10 m/s
After the man gets into the boat, the total mass becomes 500+50 = 550 Kg.=${\mathrm{m}}_2$
Then applying the formula we have
500×10=550×${\mathrm{v}}_2$
${\mathrm{v}}_2$​=$\frac{100}{11}$ m/s