A Dodge car of mass 1103 kg traveling to the right at 22.40 m/s collides head-on

Question

A Dodge car of mass 1103 kg traveling to the right at 22.40 m/s collides head-on with a Ford of mass 1258 kg traveling at the same speed but in the opposite direction. What is the speed of the Dodge the instant after the collision, assuming the collision is perfectly inelastic?

In which direction will the Dodge travel the instant after the collision?

Suppose now that the Dodge instead had a mass of 1413 kg (but the same speed as before). Calculate its final speed the moment after the collision, again assuming a perfectly inelastic collision.

In which direction will the Dodge travel the instant after the collision?

A popular newspaper columnist* once fielded the question from one of her readers: Would it be better to collide head-on with an identical car traveling at the same speed or collide with a stationary brick wall? The author answered: The brick wall. Using similar calcuations as above and examining your results above, how would you respond? (Assume that the collision time is the same in both instances.)

A.The author is wrong; it is better to hit the incoming car in a head-on collision.

B.The author is right; it is better to hit the brick wall.

C.The author is wrong; it makes no difference.

Some public awareness ads have stated that being involved in a head-on collision with both cars traveling at 30 mph is like hitting a brick wall at 60 mph. How would you respond?

A.This claim is false; hitting the brick wall at 60 mph is much worse.

B.This claim is false; the head-on collision at 30 mph is much worse.

C.This claim is reasonable.

Explanation / Answer

Applying momentum conservation,

Pi = Pf

(1103 x 22.40) + (1258 x -22.40) = (1103 + 1258) v

v = - 1.47 m/s

- Dodge will travel in opposite direction. .....Ans

now (1413 x 22.40) + (1258 x - 22.40) = (1103 + 1413) v

v = 1.38 m/s

- Dodge will travel in same direction as it was travelling. ....Ans

--------------------

for moving cars: (m v) + (- m v) = (m + m) vf

vf = 0

Imulse= m v

in case of stationary wall, Impulse = m v { as vf = 0}

Ans: (c) The author is wong,; it makes no difference.

impulse = m (30 - 0 )

for wall , = m (60 - 0)

in case of wall, impulse is double as in case of cars.

Ans: (A) claim is false, hitting a brick wall is much worse.

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