Question

A 9.13e+3 kg railroad car is rolling at 3.15 m/s when a 4.20e+3 kg load of gravel is suddenly dropped in from directly above. What is the car's speed immediately after the gravel is dropped in?

Answer 1

Answer:

6.85 m/s

Explanation:

We can solve the problem by using the law of conservation of momentum.

In fact, since there are no external forces acting, the total momentum before and after must be conserved. So we can write:

$m_1 v_1 = m_2 v_2$

where

$m_1 = 9.13\cdot 10^3 kg$ is the initial mass of the car

$v_1 = 3.15 m/s$ is the initial speed of the car

$m_2 = 9.13\cdot 10^3 kg - 4.20\cdot 10^3 kg=4.93\cdot 10^3 kg$ is the mass of the car after the load of gravel is dropped

v2 is the final speed of the car

Solving for v2, we find

$v_2 = \frac{m_1 v_1}{m_2}=\frac{(9.13\cdot 10^3)(3.15 m/s)}{4.93\cdot 10^3}=6.85 m/s$