Question

Two pieces of clay are moving directly toward each other. When they collide, they stick together and move as one piece. One piece has a mass of 324 grams and is moving to the right at a speed of 1.15 m/s. The other piece has mass 625 grams and is moving to the left at a speed of 0.87 m/s. What fraction of the total initial kinetic energy is lost during the collision? In other words what is (KE?i???KE?f???? )/ KEi?

2. A 1510 kg mid-size car traveling east at 16.4 m/s collides with a 2320 kg mid-size truck traveling north at 31 m/s. Assume that the vehicles upon collision lock together in such a way as to prevent them from separating or rotating. Calculate the angle northward from due east (counterclockwise) that the locked cars move toward immediately after the impact.

Answer 1

One question at a time please....Please follow the rules

1) This is a case of inelastic collision

Using conservation of momentum

m1v1 + m2v2 = (m1 + m2 ) vfinal

0.324*1.15 + 0.625(-0.87) = (0.324 + 0.625)vfinal

0.3726 - 0.54375 = 0.949 vfinal

vfinal = - 0.18035 m/s

vfinal = 0.18035 m/s ( to the left)

Now, total initial kinetic energy

K.E (initial) = 1/2*m1v1² + 1/2*m2*v2²

K.E (initial) = 0.214245 + 0.23653 = 0.450776 J

K.E (final) 1/2*m1+m2*vfinal²

K.E (final) = 0.015433 J

so, Kinetic energy lost = (0.450776 - 0.015433) / 0.450776

Kinetic energy lost = 0.435343 / 0.450776

Kinetic energy lost = 0.96576 ( or 96.57 %)