Question

Three cars are approaching an intersection. Car #1 has mass 1500 kg and is driving East at 10 m/s. Car #2 has mass 2500 kg and is driving North at 20 m/s. Car #3 has mass 2000 kg and is driving South at 30 m/s.

a) If the three cars collide simultaneously and stick together, what is their final velocity (magnitude and direction)?

b) What fraction of the initial kinetic energy is lost in the collision?

c) If the collision last 0.5 seconds, what is the magnitude of the average force exerted on car #1? You may neglect any effects of friction between the cars' tires and the road during the collision.

d) If the coefficient of kinetic friction between the cars' tires and the road is uk=0.3, how far do the cars skid after the collision, before coming to rest?

please show the steps on how to solve each step.

Answer 1

a] Use conservation of momentum for both horizontal and vertical directions.

for horizontal direction:

1500(10) + 0 + 0 = (1500 + 2500 + 2000)V_x

=> V_x = 0.250 m/s

in vertical direction

0 + 2500(20) - 2000(30) = (6000)V_x

V_y = 1.667 m/s

so, the magnitude of the final velocity is:

and direction counter-clockwise from positive x axis (from east) is:

.

b] Fraction of energy lost =

Where

and

substitute the values to get the answer as a fraction.

c] Magnitude of average force on the car = change in momentum of car 1 / time

=> F = 1500(10 - 0.25)/0.5 = 29250 N.

d] The acceleration (deceleration) due to friction is:

a = - u_kg = - 0.3(9.8) = - 2.94 m/s².

now use

V_f = 0

so, S = 0.483 m.

This is how far the cars travel after the collision.