A 2700-kg car traveling at 30 m/s collides with a 1600-kg car at rest at a stop sign. If the 2700-kg car has a velocity of 7.5 m/s after the collision, find the velocity of the 1600-kg car after the collision.

Explanation:

There are generally two types of collisions between objects - elastic and inelastic.

Elastic collisions are those that converse kinetic energy. Inelastic are those that do not conserve kinetic energy.

In the ideal inelastic collision and elastic collisions, momentum is conserved.

Typically, ideal inelastic collisions are represented when both masses stick together after the collision.

The problem statement gives no indication that this is an ideal inelastic collision (the cars stick together) or an inelastic collision (no energy degradation expression is given). Therefore, we should assume that the cars are experiencing an elastic collision.

Since both momentum and kinetic energy are converved, we can observe that...

where v is the initial velocity and u is the final velocity (after the collision)

The problem statement gives us three of the four unknowns. So we can easily apply either equation to solve the the velocity of the 1600-kg car after the collision. Momentum is easier to work with.

most voltage drop calculators want the distance between the source and destination. then internally, they double it so both the source and return wires are calculated. but you need to know which distance the calculator uses so you're not half of what you should be or double what you should be.

most circuit breakers will not take a #6 directly, so the electrician will have to splice on a smaller pigtail. you also have to upsize the grounding wire when you upsized the ungrounded (hot) wire (see note in 250.122). this will preclude using most cable assemblies, as a cable with #6 copper will most likely only have a #10 grounding conductor. you need a #6 grounding conductor if the circuit is 20a. a #8 cable is no better, as it will have a #10 grounding conductor too. but most 20a breakers will take a #8 directly.

in your related post, you didn't say what type of load this is. if it is a light or heater, voltage drop won't be a big issue. if this is a motor, the current needs are huge at startup and the voltage drop can pose a problem. if this is a motor, knowing if the motor is under much load during startup will answer the voltage drop effect (a saw is easier to start than a compressor).