A 710-kgcar stopped at an intersection is rear-ended by a 1780-kgtruck moving wi

Question

A 710-kgcar stopped at an intersection is rear-ended by a 1780-kgtruck moving with a speed of 14.0m/s. A) If the car was in neutral and its brakes were off, so that the collision is approximately elastic, find the final speed of the truck? B) Find the final speed of the car. A 710-kgcar stopped at an intersection is rear-ended by a 1780-kgtruck moving with a speed of 14.0m/s. A) If the car was in neutral and its brakes were off, so that the collision is approximately elastic, find the final speed of the truck? B) Find the final speed of the car. A) If the car was in neutral and its brakes were off, so that the collision is approximately elastic, find the final speed of the truck? B) Find the final speed of the car.

Explanation / Answer

we use both conservation of momentum and energy since the collision is assumed elastic conservation of momentum gives us: 1780kg x 14 m/s = 1780v1 + 720 v2 where v1 and v2 are the velocities fo the truck and car after collision we can write this as v1=(24360-710v2)/1780 (eq.1) energy conservation gives us: 1/2(1780)(14)^2=1/2(17480)(v1)^2 +1/2(710)(v2)^2 divide through by 1/2(1780): 14^2=v1^2+0.51v2^2 or 196=v1^2+0.51v2^2 (eq.2) substitute eq. 1 into eq. 2 v1=(24360-710v2)/1740 196=(24360-710v2)^2/1740^2 +0.43v2^2 this is a quadratic in v2; expand the right hand side and solve for v2 via the quadratic equation this yields v2=19.6m/s and v1=5.55 m/s both vehicles continue to move forward after the collision

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