A car moving at 300 m/s has a collision and comes to rest in 0.1 seconds. The dr

Question

A car moving at 300 m/s has a collision and comes to rest in 0.1 seconds. The driver is brought to rest by an air bag in 0.13s. The driver has a mass of 50 kg.

1) what average force is exerted by the airbag on the driver?

2) calculate the work done on the driver by the air bag.

3) How far does the driver travel while stopping?

4) Two suitcases fly off the roof of the car moving at 12 m/s relative to the ground and the lid of one pops open, puching them apart. The opened suitcase, with a mass of 20 kg, is moving at 6 m/s after they seperate.If the second suitcase has a mass of 15 kg, what is its velocity after they separate? Is this a perfectly elastic interaction?

5) If the suitcases are 2.0 meters abouve the ground when they separate, how far from the point of separation does each of them hit the ground?

Explanation / Answer

1) initial velocity of car and driver, u = 300 m/s

acceleration of driver, a_driver = (v-u)/t

= (0-300)/0.13

= -2307.7 m/s^2

Force on driver = m_driver*a_driver

= 50*2307.7

= 115385 N

2) Workdone by air bag on the driver = chnage in kinetic enrgy of the driver

= 0.5*m*(v^2-u^2)

= 0.5*50*(0^2 - 300^2)

= -2.25*10^6 J

3) s = u*t + 0.5*a*t^2

= 300*0.13 - 0.5*2307.7*0.13^2

= 19.5 m

4)
aplly momentum conservation

(m1+m2)*u = m1*v1 + m2*v2

v2 = (m1+m2)*u/m2 - m1*v1/m2

= (20+15)*12/15 - 20*6/15

= 20 m/s

5) let t is the time taken to reach the ground.

H = uy*t + 0.5*g*t^2

H = 0 + 0.5*g*t^2

t = sqrt(2*H/g)

= sqrt(2*2/9.8)

= 0.639 s

so, x1 = v1*t = 6*0.639 = 3.83 m

x2 = v2*t = 20*0.639 = 12.7 m

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