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.

Explanation / Answer

given: m1=1103kg, u1=22.40 m/s

m2=1258 kg, u=-22.40 m/s

In a fully inelastic collision the two objects stick together and move as one combined object after the collision. So the final speed of both vehicles will be the same and momentum (but not kinetic energy) is conserved.
(a) Taking the +ve direction to the right:
Initial momentum =(m1*u1+m2*u2)= (1103*22.4) + (1258*(-22.4)) = -3472 kg.m/s
Final momentum = (m1+m2)*Vf =(1103+1258) * Vf = -3472 kg.m/s
so Vf = -3472 / 2361 = -1.47 m/s
(to the left, in the direction of the heavier vehicle)

(b)m1=1413 kg

Taking the +ve direction to the right:
Initial momentum = (m1*u1+m2*u2)=(1413*22.4) + (1258*-22.4) = +3472 kg.m/s
Final momentum = (m1+m2)*Vf=(1413+1258) * Vf = +3472 kg.m/s
so Vf = +3472 / 2361 = +1.30 m/s
(to the right, again in the direction of the heavier vehicle)

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