Starting with an initial speed of 3.05 m/s at a height of 0.440 m, a 2.56-kg bal

Question

Starting with an initial speed of 3.05 m/s at a height of 0.440 m, a 2.56-kg ball swings downward and strikes a 5.14-kg ball that is at rest, as the drawing shows. (a) Using the principle of conservation of mechanical energy, find the speed of the 2.56-kg ball just before impact. (b) Assuming that the collision is elastic, find the velocity (magnitude and direction) of the 2.56-kg ball just after the collision. (c) Assuming that the collision is elastic, find the velocity (magnitude and direction) of the 5.14-kg ball just after the collision. (d) How high does the 2.56-kg ball swing after the collision, ignoring air resistance? (e) How high does the 5.14-kg ball swing after the collision, ignoring air resistance?

Explanation / Answer

v=sqrt(2gh)=sqrt(2*9.81*0.44)=2.93 m/s a)2.93 m/s b)v1`=(m1-m2)*v1/(m1+m2) v1`=(2.56-5.14)*2.93/(2.56+5.14)=-0.98 m/s direction is towards left magnitude=0.98 m/s c)v2`=2*m1*v1/(m1+m2)=1.94 m/s direction is towards right magnitude=1.94 m/s d)h=v1`^2/2g=0.0489 m e)h=v2`^2/2g=0.191 m

Related Questions

Navigate

• Browse (All)
• Browse S
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.