Question

1. a 4.50 x10³ kg railway car moving east at a velocity of 5.0 m/s on a level frictionless track when it collides with a stationary 6.50 x10³ kg caboose. if the two cars lock together upon impact, how fast are they moving after the collision?
2. a 2.00 kg rock is dropped from the top of a 30.0 m high building. calculate the balls momentum at the time that it strikes the ground.

Explanation:

Problem 3

Step1: Apply conservation of momentum

$$m_1v_1 + m_2v_2=(m_1 + m_2)v$$
Here, \(m_1 = 4.50\times10^{3}\space kg\), \(v_1=5.0\space m/s\), \(m_2 = 6.50\times10^{3}\space kg\), \(v_2 = 0\space m/s\)

Step2: Substitute values into the formula

$$(4.50\times 10^{3}\times5.0)+(6.50\times 10^{3}\times0)=(4.50\times 10^{3}+6.50\times 10^{3})v$$
$$22500=(11000)v$$

Step3: Solve for \(v\)

$$v=\frac{22500}{11000}\approx 2.05\space m/s$$

Step1: Find the final velocity using kinematic equation

$$v^{2}=u^{2}+2gh$$
Since \(u = 0\space m/s\) (dropped), \(g = 9.8\space m/s^{2}\), \(h = 30.0\space m\)
$$v=\sqrt{2gh}=\sqrt{2\times9.8\times30.0}=\sqrt{588}\approx 24.25\space m/s$$

Step2: Calculate momentum

Momentum \(p=mv\), where \(m = 2.00\space kg\), \(v\approx24.25\space m/s\)
$$p=2.00\times24.25 = 48.5\space kg\cdot m/s$$

Answer:

The velocity of the combined cars after the collision is approximately \(2.05\space m/s\)

Problem 4