A 12,000-lb bus collides with a 2800-lb car. The velocity of the bus before the collision is vB = 18i (ft/s) and the velocity of the car is vC = 33j (ft/s). The two vehicles become entangled and remain together after the collision. The coefficient of kinetic friction between the vehicles' tires and the road is μk = 0.6. Find (a) velocity of combined center of mass immediately after the collision, (b) final position (assume brakes are on and you have skidding, not rolling). (c) If the collision lasted 0.5 seconds, what was the impulsive force of the bus on the car?

Question

Physics

Eddie Orozco

6 March, 06:40

A 12,000-lb bus collides with a 2800-lb car. The velocity of the bus before the collision is vB = 18i (ft/s) and the velocity of the car is vC = 33j (ft/s). The two vehicles become entangled and remain together after the collision. The coefficient of kinetic friction between the vehicles' tires and the road is μk = 0.6. Find (a) velocity of combined center of mass immediately after the collision, (b) final position (assume brakes are on and you have skidding, not rolling). (c) If the collision lasted 0.5 seconds, what was the impulsive force of the bus on the car?

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Answers (1)

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William Rollins · Answer 1

(a) 20.84 ft/s

(b) 11.24 ft

(c) - 68160 N

Explanation:

Parameters given:

Mass of Bus, Mb = 12000 lb

Mass of car, Mc = 2800 lb

Initial speed of bus before collision, u (b) = 18 ft/s

Initial speed of car before collision, u (c) = 33 ft/s

Coefficient of friction, μk = 0.6

(a) Combined velocity after collision.

Since the bus and car are entangled and move together after the collision, they have the same velocity after collision.

Using the law of conservation of momentum, we have:

Mb*u (b) + Mc*u (c) = Mb*v (b) + Mc*v (c)

Where v (b) = velocity of the bus, L after collision,

v (c) = velocity of the car after collision.

Since v (b) = v (c) = v,

Mb*u (b) + Mc*u (c) = (Mb + Mc) * v

=> (12000 * 18) + (2800 * 33) = (12000 + 2800) * v

=> 308400 = 14800 * v

=> v = 20.84 ft/s

Combined velocity after collision is 20.84 ft/s

(b) The force acting on the bus and car after collision is given as:

F = m*a

Where F = force,

m = combined mass of bus and car,

a = acceleration of the bus and car after collision.

We know that the only force acting on the combined mass of the bus and car is the Frictional force, Fr and it is given as:

Fr = μk*m*g

Where g = acceleration due to gravity

Hence,

Fr = ma

=> - μk*m*g = m*a

The negative sign signifies that the fictional force is acting in the opposite direction to the lotion.

=> a = - μk*g

a = - 0.6 * 32.2

a = - 19.32 ft/s^2

Using one of the equations of linear motion, we can find the distance moved by the car and bus after collision:

v*v = u*u + 2*a*s

Where s = distance moved

The final velocity, v, of the car and bus is 0 because they come to rest and the initial velocity, u, is 20.84 ft/s, the velocity of the car and bus after collision.

Hence,

0 = (20.84*20.84) + (-2*19.32*s)

=> s = - 434.3056/-38.64

s = 11.24 ft

(c) Impulsive force is the force that two bodies which are colliding exert on one another. It is given mathematically as

I. F. = (momentum change) / time

Momentum change of the bus is:

Momentum change = final momentum - initial momentum

Momentum change = Mb*v (b) - Mb*u (b)

Momentum change = (12000*20.84) - (12000*18)

Momentum change = 250080 - 216000

Momentum change = - 34080 kgft/s

=> I. F. = - 34080/0.5

I. F. = - 68160 N

The Impulsive force of the bus on the car is - 68160N

The negative sign means the force is acting opposite to the motion of the car.