A car of mass 1800 kg stopped at a traffic light is struck from the rear by a 90

Question

A car of mass 1800 kg stopped at a traffic light is struck from the rear by a 900 kg car, and the two become entangled. a) If the smaller car was moving at 20.0 m/s before the collision, what is the velocity of the entangled cars after the collision? b) What would be the final speed if the two cars each had a mass of 900 kg? A 0.82 kg ball traveling to the right at 0.50 m/s collides head on with a 0.60 kg ball originally moving to the left at 0.25 m/s. If the collision is perfectly elastic, what is the velocity of each ball after the collision? (Head on means all motion takes place on a straight line). At an Amusement Park the radius of the rotor device is 3.66m. The coefficient of static friction necessary to prevent riders from falling is 0.45. Find the required speed to insure safety. During an emergency braking a car skids 300m away. The initial speed of the vehicle was 60m/s. What is the coefficient of kinetic friction?

Explanation / Answer

5A)

Before collision

smaller car weight = 900kg

smaller car velocity = 20 m/s

bigger car velocity before collision = 0m/s (rest)

a) Momentum of system(before collision) = 900*20 + 0 = 18000 kgm/s

Since the cars become entangled,

Momentum = (900+1800) * v, Momentum is conserved, so 18000 = (900+1800)*v

v= 18000/2700 = 6.67 m/s

b) 900*20 = (900+900) V {Since both cars has mass = 900kg)

V = 10 m/s

5B)

m1=0.82Kg, v1initial=0.50m/s (right)

m2=0.60kg v2initial= -0.25m/s(as it is left)

For perfectly elastic collision,

by momentum and energy conservation

v1final = (m1-m2)/(m1+m2) *v1initial + (2m2)/(m1+m2)* v2initial

= (0.22/1.42) *0.5 + (1.2/1.42) *(-0.25)

=0.077 - 0.211 = - 0.134 m/s

v2final = 2m1/(m1+m2) *v1initial + (m2-m1)/(m2+m1) * v2 initial

=(1.64/1.42)*0.5 + (0.6-0.82)/(1.42) * (-0.25)

=0.577 +0.038 = 0.615 m/s

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