Starting with an initial speed of 5.64 m/s at a height of 0.130 m, a 2.25-kg bal

Question

Starting with an initial speed of 5.64 m/s at a height of 0.130 m, a 2.25-kg ball swings downward and strikes a 4.56-kg ball that is at rest, as the drawing shows. (a) Using the principle of conservation of mechanical energy, find the speed of the 2.25-kg ball just before impact. (b) Assuming that the collision is elastic, find the velocity (magnitude and direction) of the 2.25-kg ball just after the collision. (c) Assuming that the collision is elastic, find the velocity (magnitude and direction) of the 4.56-kg ball just after the collision. (d) How high does the 2.25-kg ball swing after the collision, ignoring air resistance? (e) How high does the 4.56-kg ball swing after the collision, ignoring air resistance?

http://i63.tinypic.com/2j1991h.png

Explanation / Answer

let,

masses of the balls, m1=2.25kg, m2=4.56kg

initial speed of m1 is, uo=5.64 m/sec

initial speed of m2 is, vo=0

height, h=0.13 m

a)

by using law of conservation of enegry,

1/2*m1*u1^2=m1*g*h

==> u1=sqrt(2*g*h)

u1=sqrt(2*9.8*0.13)

u1=1.6 m/sec

before impact, speed of m1 ball is, u1=1.6 m/sec

now,

given that, collision is elastic and after impact,

b)

velocity of m1 is, vf1=(m1-m2)*u1/(m1+m2)

vf1=(2.25-4.56)*1.6/(2.25+4.56)

vf1=-0.543 m/sec ( back ward)

c)

velocity of m2 is, vf2=(2*m1)*u1/(m1+m2)

vf2=2*(2.25)*1.6/(2.25+4.56)

vf2=1.06 m/sec

d)

by using conservation of enegry,

m1*g*h1=1/2*m1*vf1^2

===> h1=vf1^2/2*g

h1=0.543^2/(2*9.8)

h1=1.504 cm

height reached by the ball m1 is, h1=1.504 cm

e)

by using conservation of enegry,

m2*g*h2=1/2*m2*vf2^2

===> h2=vf2^2/2*g

h2=1.06^2/(2*9.8)

h2=0.0573 m or 5.73cm

height reached by the ball m1 is, h2=5.73cm

Related Questions

Navigate

• Browse (All)
• Browse S
• Subjects

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