Question

1. a 6.0 x 103 kg railway car is coasting along the track at 7.0 m/s. suddenly a 2.0 x 103 kg load of coal is dumped into the car. what is its new velocity?

Explanation:

Step1: Apply conservation of momentum

The initial momentum of the system is the momentum of the railway - car only since the coal is initially at rest in the vertical - direction. The formula for momentum is $p = mv$. The initial momentum $p_i=m_1v_1$, where $m_1 = 6.0\times10^{3}\text{ kg}$ and $v_1 = 7.0\text{ m/s}$. So, $p_i=(6.0\times10^{3}\text{ kg})\times(7.0\text{ m/s}) = 4.2\times10^{4}\text{ kg}\cdot\text{m/s}$.

Step2: Calculate the final mass

After the coal is dumped into the car, the final mass of the system $m_f=m_1 + m_2$, where $m_2=2.0\times10^{3}\text{ kg}$ and $m_1 = 6.0\times10^{3}\text{ kg}$. So, $m_f=(6.0\times10^{3}\text{ kg})+(2.0\times10^{3}\text{ kg})=8.0\times10^{3}\text{ kg}$.

Step3: Use conservation of momentum to find final velocity

According to the law of conservation of momentum $p_i = p_f$, and $p_f=m_fv_f$. Since $p_i = p_f$, we have $v_f=\frac{p_i}{m_f}$. Substituting $p_i = 4.2\times10^{4}\text{ kg}\cdot\text{m/s}$ and $m_f = 8.0\times10^{3}\text{ kg}$ into the formula, we get $v_f=\frac{4.2\times10^{4}\text{ kg}\cdot\text{m/s}}{8.0\times10^{3}\text{ kg}}=5.25\text{ m/s}$.

Answer:

$5.25\text{ m/s}$