Question

A 500 kg boat has an initial speed of $10m{s}^{-1}$ 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 attains a common speed is:

• A. $\frac{100}{11}m{s}^{-1}$
• B. $\frac{10}{11}m{s}^{-1}$
• C. $\frac{50}{11}m{s}^{-1}$
• D. $\frac{5}{11}m{s}^{-1}$

Answer 1

The correct option is A $\frac{100}{11}m{s}^{-1}$

We know from the law of conservation of momentum
$m_1{v}_1=m_2{v}_2$
Here $m_1$=500 Kg.
$v_1$=10 m/s
After man gets into the boat, the total mass becomes 500+50 = 550 Kg.= $m_2$
Then applying the formula we have
$500\times 10=550\times v_2$
$v_2=\frac{100}{11}$ m/s