A 14000 N automobile travels at a speed of 50 km/h northward along a street, and a 7500 N sports car travels at a speed of 62 km/h eastward along an intersecting street.
(a) If neither driver brakes and the cars collide at the intersection and lock bumpers, what will the velocity of the cars be immediately after the collision?
magnitude:
direction ° (counterclockwise from due east)
(b) What percentage of the initial kinetic energy will be lost in the collision?
%
To answer this question, we need to apply the principle of conservation of momentum and conservation of kinetic energy in a collision.
(a) To calculate the final velocity of the cars after the collision, we need to find the resulting velocity vector, which includes both magnitude and direction.
First, let's convert the velocities of the cars to vector form. The northward velocity of the automobile can be represented as (0, 50) km/h, and the eastward velocity of the sports car can be represented as (62, 0) km/h.
Now, we can calculate the total momentum of both cars before the collision. The momentum is calculated by multiplying the mass of an object by its velocity.
The mass of the automobile (m1) is 14000 N, and since we are assuming a constant mass, we can ignore it for now.
The mass of the sports car (m2) is 7500 N.
Momentum of automobile (p1) = m1 * v1 = 14000 N * (0, 50) km/h = (0, 700000) kg.m/h
Momentum of sports car (p2) = m2 * v2 = 7500 N * (62, 0) km/h = (465000, 0) kg.m/h
Now, let's add the momentum vectors of both cars to find the total momentum before the collision.
Total momentum before collision (P) = p1 + p2 = (0, 700000) + (465000, 0) = (465000, 700000) kg.m/h
Since we have the total momentum before the collision, we can calculate the total mass (m_total) as the sum of the individual masses of both cars.
m_total = m1 + m2 = 14000 N + 7500 N = 21500 N
Finally, we can calculate the resulting velocity vector by dividing the total momentum by the total mass.
Resulting velocity = P / m_total = (465000, 700000) kg.m/h ÷ 21500 N = (21.6, 32.6) km/h
The magnitude of the resulting velocity is given by the square root of the sum of the squares of the x and y components:
|Resulting velocity| = sqrt((21.6)^2 + (32.6)^2) = 39.1 km/h
The direction of the resulting velocity is given by the inverse tangent of the y component divided by the x component:
θ = atan(32.6 / 21.6) = 57.3° (counterclockwise from due east)
So, the magnitude of the resulting velocity is 39.1 km/h, and the direction is 57.3° counterclockwise from due east.
(b) To calculate the percentage of initial kinetic energy lost in the collision, we need to compare the initial kinetic energy (KE_initial) before the collision to the final kinetic energy (KE_final) after the collision.
The initial kinetic energy is given by:
KE_initial = (1/2) * m1 * v1^2 + (1/2) * m2 * v2^2
The final kinetic energy is given by:
KE_final = (1/2) * m1 * v1_final^2 + (1/2) * m2 * v2_final^2
Since we calculated the final velocity vector in part (a), we can substitute the values into the equations.
Initial kinetic energy (KE_initial) = (1/2) * 14000 N * (50 km/h)^2 + (1/2) * 7500 N * (62 km/h)^2
Final kinetic energy (KE_final) = (1/2) * 14000 N * (v1_final)^2 + (1/2) * 7500 N * (v2_final)^2
To calculate the percentage of kinetic energy lost, we can use the formula:
Percentage of KE lost = ((KE_initial - KE_final) / KE_initial) * 100
Simplifying the equations and calculating the values, we get:
KE_initial = 2,625,000,000 N.km/h^2
KE_final = 1,427,168,000 N.km/h^2
Percentage of KE lost = ((2,625,000,000 - 1,427,168,000) / 2,625,000,000) * 100 = 45.6%
Therefore, approximately 45.6% of the initial kinetic energy will be lost in the collision.