Question

Batman (mass=100 kg) jumps straight down from a bridge in to a boat (mass = 580 kg) in which a criminal is fleeing. The velocity of the boat is initially +14 m/s. What is the velocity of the boat after Batman lands in it?

Answer 1

Answer:

$v_b=11.94\ m/s$

Step-by-step explanation:

Linear Momentum

The momentum of a system of masses m1,m2,m3... with speeds v1,v2,v3,... is given by

$M=m_1v_1+m_2v_2+m_3v_3+...$

When some interactions take place into the system of masses with no external forces interferring, the total momentum is not changed, which means that if the new speeds are v1', v2', v3'... then:

$m_1v_1+m_2v_2+m_3v_3+...=m_1v_1'+m_2v_2'+m_3v_3'+...$

The problem relates the story of Batman (m1=100kg) who is initially assumed at rest and lands on a boat with mass m2=580 kg initially moving at 14 m/s to the positive reference. When the collision takes place, both masses join and a common speed $v_b$ is achieved by the common mass. Applying the conservation of momentum, we have

$m_1v_1+m_2v_2=(m_1+m_2)v_b$

Since $v_1=0$

$m_2v_2=(m_1+m_2)v_b$

Solving for $v_b$

$\displaystyle v_b=\frac{m_2v_2}{m_1+m_2}$

$\displaystyle v_b=\frac{(580)(14)}{100+580}$

$\boxed{v_b=11.94\ m/s}$