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## Homework Statement

A 0.26 kg cue ball with a velocity 1.2 m/s collides elastically with a 0.15 kg billiard ball at rest.

What percentage of the initial kinetic energy is transferred to the billiard?

m1= 0.26kg

Vi1= 1.2 m/s

Vf1= ?

m2= 0.15kg

Vi2= 0 m/s

Vf2= ?

## Homework Equations

(m1v2+m2v2)i=(m1v2+m2v2)f

ΔKE= ((.5)(m1)(Vi1)

^{2}+(.5)(m2)(Vi2)

^{2})i -((.5)(m1)(Vf1)

^{2}+(.5)(m2)(Vf2)

^{2})f

%ΔKE= 100(F-I/I)

## The Attempt at a Solution

(0.26kg)(1.2m/s)+0=(0.26kg)(-vf1)+(0.15kg)(-vf2)

(0.312 kg m/s)=(0.26kg)(-vf1)+(0.15kg)(-vf2)

(0.312 kg m/s)-(0.26kg)(-vf1)= (0.15kg)(-vf2)

(0.312 kg m/s)+(0.26kg)(vf1)= (-0.15kg)(vf2)

((0.312 kg m/s)+(0.26kg)(vf1))/(-0.15kg)=(vf2)

(0.312 kg m/s)=(0.26kg)(-vf1)+(0.15kg)[[(0.312 kg m/s)+(0.26kg)(vf1)]/(-0.15kg)]

(0.312 kg m/s)=(0.26kg)(-vf1)-[(0.312 kg m/s)]-[(0.26kg)(vf1)]

(0.624 kg m/s)=(0.26kg)(-vf1)-[(0.26kg)(vf1)]

(0.624 kg m/s)=(-0.52vf1)

vf1= -1.2 m/s

((0.312 kg m/s)+(0.26kg)(-1.2 m/s))/(-0.15kg)=(vf2)

((0.312 kg m/s)+(-0.312 kg m/s))/(-0.15kg)=(vf2)

0=vf2

ΔKE= ((.5)(0.26kg)(1.2 m/s)

^{2}+(.5)(0.15kg)(0 m/s)

^{2})i -((.5)(0.26kg)(Vf1)

^{2}+(.5)(0.15kg)(Vf2)

^{2})f

The answer is suppose to be 93%

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