$A$ man weighing $60 \, kg$ runs along the rails with a velocity of $18 \, km \, h^{-1}$ and jumps into a car of mass $1 \, \text{quintal} \, (100 \, kg)$ standing on the rails. Calculate the velocity with which the car will start travelling along the rails.
Mass of the man,$m_{1} = 60 \, kg$.
Initial speed of the man,$u_{1} = 18 \, km \, h^{-1} = \frac{18 \times 1000}{3600} \, m \, s^{-1} = 5 \, m \, s^{-1}$.
Mass of the car,$m_{2} = 100 \, kg$.
Initial speed of the car,$u_{2} = 0 \, m \, s^{-1}$.
Let the final velocity of the car with the man be $v$.
According to the law of conservation of momentum:
$m_{1}u_{1} + m_{2}u_{2} = (m_{1} + m_{2})v$.
Substituting the values:
$(60 \times 5) + (100 \times 0) = (60 + 100)v$.
$300 = 160v$.
$v = \frac{300}{160} \, m \, s^{-1} = 1.875 \, m \, s^{-1}$.
Converting to $km \, h^{-1}$:
$v = 1.875 \times \frac{18}{5} \, km \, h^{-1} = 6.75 \, km \, h^{-1}$.
Initial speed of the man,$u_{1} = 18 \, km \, h^{-1} = \frac{18 \times 1000}{3600} \, m \, s^{-1} = 5 \, m \, s^{-1}$.
Mass of the car,$m_{2} = 100 \, kg$.
Initial speed of the car,$u_{2} = 0 \, m \, s^{-1}$.
Let the final velocity of the car with the man be $v$.
According to the law of conservation of momentum:
$m_{1}u_{1} + m_{2}u_{2} = (m_{1} + m_{2})v$.
Substituting the values:
$(60 \times 5) + (100 \times 0) = (60 + 100)v$.
$300 = 160v$.
$v = \frac{300}{160} \, m \, s^{-1} = 1.875 \, m \, s^{-1}$.
Converting to $km \, h^{-1}$:
$v = 1.875 \times \frac{18}{5} \, km \, h^{-1} = 6.75 \, km \, h^{-1}$.
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