Question

5. (3 pts) A 1200-kg Toyota Camry and a 2200-kg Lexus collide at right angles in an intersection. They lock together and skid 22m; the coefficient of friction between their (nonrolling) tires and the roadway is measured to be 0.91. Assume that the collision happens very quickly, followed by the skid. Therefore the impulse delivered by the road to the cars during the collision is negligible, but not the impulse delivered during the skid. Show that at least one car must have exceeded the 25 km/hr speed limit.

Answer 1

for random value of angle, one of initial velocities is > 21.6 m/s which is much above speed limit of 6.9 m/s(25 km/h)

Step-by-step explanation:

Force of friction causes deceleration,

`a = (f)/(m) = (μN)/(m)`

`= (μ(m1 + m2)g)/((m1+m2)) = μg`

`=0.91* 9.8 =  8.918 m/s2`

If v is velocity,

after collision, final velocity, vf = 0;

Applying,
`vf^2 - vi^2 = 2a.S</p><p>`

`v^2 = 2a.S = 2* 8.918* 22 = 392.4</p><p>`

v = 19.81 m/s

Let first car moving along x axis and 2nd car moving along Y axis just before collision.

considering θ be angle of direction with x-axis of motion after collision.

Let v_1 and v2 be the velocities of 1st & 2nd cars before collision.

By using conservation of momentum along the x axis;

`1200.v1 = 3400.vx = 3400* 19.81 cos\theta`

`2200.v2 = 3400.vy =3400* 19.81 sin \theta`

Hence, v1 = 56.13 cos θ

v2 = 30.62 sin θ

considering
`\theta = 45 degree`

then v1 = 39.7 /s

v2 = 21.6 /s

The speed limit of 25 km/h = 6.9 m/s

For θ > 45 cos θ < 0.707; v1 < 39.7 m/s

But sin θ > 0.707; v2> 21.6 m/s

For θ < 45 cos θ > 0.707; v1 > 39.7 m/s

But sin θ < 0.707;v2 < 21.6 m/s

So, for random value of angle, one of initial velocities is > 21.6 m/s which is much above speed limit of 6.9 m/s(25 km/h).