Two cars crash into each other at an intersection on a very icy road. One 3,000 lb car is traveling 45 mph NW. The other, weighing 2,000 lb, is traveling 30 mph E. After the collision, their bumpers lock together. What is the velocity of the wreck as it slides off the road ?
The final velocity equals the total momentum (which stays the same) divided by the combined mass (5000 lb/g or 155.3 slugs)
Add, as vectors, the two initial momenta of the two cars to get the total momentum.
In this case you can keep the speeds in mph and the masses in lbm (pounds-mass), to get a final result in mph. In general it is safest, when working in British-American units, to convert speed to ft/s and mass to slugs.
To determine the velocity of the wreck as it slides off the road after the collision, we need to analyze the momentum of the two cars before and after the collision. The law of conservation of momentum states that the total momentum before the collision is equal to the total momentum after the collision, assuming no external forces act on the system.
First, we need to calculate the momentum of each car before the collision. Momentum (p) is calculated by multiplying the mass (m) of an object by its velocity (v), using the formula p = m * v.
For the first car:
Mass (m1) = 3,000 lb
Velocity (v1) = 45 mph NW (Northwest)
Since the velocity is given in mph, we need to convert it to ft/s (feet per second) to ensure consistent units. To convert mph to ft/s, we multiply by the conversion factor 1.47 (1 mph = 1.47 ft/s).
Converted velocity (v1) = 45 mph * 1.47 ft/s(mph-ft/s)
≈ 66.15 ft/s NW
Momentum of the first car (p1) = m1 * v1
= 3,000 lb * 66.15 ft/s NW
= 197,930 lb-ft/s NW
Similarly, for the second car:
Mass (m2) = 2,000 lb
Velocity (v2) = 30 mph E (East)
Converted velocity (v2) = 30 mph * 1.47 ft/s(mph-ft/s)
≈ 44.1 ft/s E
Momentum of the second car (p2) = m2 * v2
= 2,000 lb * 44.1 ft/s E
= 88,200 lb-ft/s E
Now, let's determine the direction of the combined momentum after the collision. When the two cars collide, their bumpers lock together, and they move as one unit. Since one car is traveling northwest (NW) and the other east (E), the resulting direction will be a combination of the two.
We can calculate the combined momentum by adding the momentum vectors of the two cars. To do this, we need to consider that the east direction is positive, and the west direction is negative.
Combined momentum = p1 + p2
In terms of magnitude:
Combined momentum = |p1| + |p2|
And in terms of direction:
Combined momentum direction = tan^(-1)(|(v1-v2)/(v1+v2)|)
Let's calculate the combined momentum:
Combined momentum = √(197,930^2 + 88,200^2)
= √(39,174,904,900 + 7,774,440,000)
≈ √46,949,344,900
≈ 216,462 lb-ft/s
Combined momentum direction = tan^(-1)((66.15 - 44.1) / (66.15 + 44.1))
Examples of simplifying this further:
= tan^(-1)(22.05 / 110.25)
= tan^(-1)(0.2)
≈ 0.197 radians (approx.)
Now that we have the magnitude and direction of the combined momentum, we can convert them to a resultant velocity.
Resultant velocity magnitude = √(combined momentum^2 / (m1 + m2)^2)
= √(216,462^2 / (3,000 + 2,000)^2)
= √(46,854,238,244 / 25,000,000)
≈ √1.874,169,529.76
≈ 43,311.8 ft/s
Resultant velocity direction = combined momentum direction
Hence, the velocity of the wreck as it slides off the road is approximately 43,311.8 ft/s in the direction of 0.197 radians (East of North).