Question

A truck with a mass of 1730 kg and moving with a speed of 14.0 m/s rear-ends a 743-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.

vcar = ____m/s

vtruck = ____m/s

Answer 1

conservation of momentum:

1730kg x 14 m/s = 1730v1 + 743 v2 where v1 and v2 are the velocities fo the truck and car after collision

v1=(24220-743v2)/1730 (eq.1)

energy conservation gives us:

1/2(1730)(14)^2=1/2(1730)(v1)^2 +1/2(743)(v2)^2

divide through by 1/2(1730):

14^2=v1^2+0.429v2^2 or

196=v1^2+0.429v2^2 (eq.2)

substitute eq. 1 into eq. 2

v1=(24220-743v2)/1730

196=(24220-743v2)^2/1730^2 +0.429v2^2

this is a quadratic in v2; expand the right hand side and solve for v2 via the quadratic equation