Question

A truck with a mass of 1420 kg and moving with a speed of 12.0 m/s rear-ends a 649 kg car stopped at an intersection. The collision is approximately elastic since the car is in neutral, the brakes are off, the metal bumpers line up well and do not get damaged. Find the speed of both vehicles after the collision in meters per second.

*v*_{car} = _______ m/s

*v*_{truck} = _______m/s

Answer #1

Now it is elastic collision, so using momentum conservation

Pi = Pf

m1u1 + m2u2 = m1v1f + m2v2f

u2 = 0 = intial speed of car

u1 = v = speed of truck

m1v = m1v1f + m2v2f

another condition for elastic collision

u1 - u2 = v2f - v1f

u2 = 0

v2f = v + v1f

m1v = m1v1f + m2v2f

from above two equations

m1v = m1*v1f + m2*(v + v1f)

v1f*(m1 + m2) = (m1 - m2)v

Solving above equation

**v1f = v*(m1 - m2)/(m1 + m2)**

**v1f = 12*(1420 - 649)/(1420 + 649) = 4.47 m/sec = final
speed of car**

v2f = v + v1f

v2f = v + v*(m1 - m2)/(m1 + m2)

**v2f = 2*m1*v/(m1+ m2)**

**v2f = 2*1420*12/(1420 + 649) = 16.47 m/sec = final speed
of truck**

**Please Upvote.**

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