The two cars collide at right angles in the intersection of two icy roads. Car A has a mass of 1965 kg and car B has a mass of 1
245 kg. The cars become entangled and move off together with a common velocity in the direction indicated. If car A was traveling 52 km/h at the instant of impact, compute the corresponding velocity of car B just before impact.
1 answer:
Answer:
U2 = 47.38m/s = initial velocity of B before impact
Explanation:
An example of the diagram is shown in the attached file because of missing angle of direction in the question
Mass A, B are mass of cars
A = 1965
B =1245
U1 = initial velocity of A = 52km/hr
U2 = initial velocity of B
V = common final velocity of two cars
BU2 = (A + B)*V sin ¤ ...eq1 y plane
AU1 = (A + B) *V cos ¤ ....equ 2plane
From equ 2
V = AU1/(A + B)*cos ¤
Substitute V into equation 1
We have
U2 = (AU1/B)tan ¤ where ¤ = angle of direction which is taken to be 30°
Substitute all parameters to get
U2 = (1965/1245)*52 * tan 30°
U2 = 47.38m/s
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The displacement is 2 m south
Explanation:
Distance and displacement are two different quantities:
- Distance is the total length of the path covered by an object during its motion, regardless of the direction. It is a scalar quantity
- Displacement is a vector connecting the initial position to the final position of motion of an object. The magnitude of the displacement is the distance in a straight line between the two points
For the car in this problem, the motion is:
10 m south
8 m north
Taking north as positive direction, we can describe the two parts of the motion as
m
Therefore, the final position of the car with respect to the original position is
which means 2 m south: so, the displacement of the car is 2 m south.
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