Question

remarks notice the balls exchanged velocities - almost as if theyd passed through each other. this is always the case when two objects of equal mass undergo an elastic head - on collision.
question is it possible to adjust the initial velocities of the balls so that both final velocities are zero? (select all that apply.)
no, because momentum conservation requires the final total momentum to be always non - zero.
no, because momentum conservation requires the final total momentum to be always zero.
no, because energy is conserved and requires the total kinetic energy to be non - zero.
no, because energy is conserved and that requires both balls to be moving after the collision.
no, because energy is conserved and that requires one ball or both to be moving after the collision.
yes, for two initial velocities equal in magnitude and opposite in direction.
practice it
use the worked example above to help you solve this problem. two billiard balls of identical mass move toward each other as shown in the figure. assume that the collision between them is perfectly elastic. if the initial velocities of the balls are ( v_{1i}=+33.1 cm/s ) and ( v_{2i}=-20.2 cm/s ), what are the velocities of the balls after the collision? assume friction and rotation are unimportant. (indicate the direction with the sign of your answer.)
( v_{1f}= cm/s )
( v_{2f}= cm/s )
exercise
hints: getting started | im stuck!
find the final velocity of the two balls if the ball with velocity ( v_{2i}=-20.2 cm/s ) has a mass equal to half that of the ball with initial velocity ( v_{1i}=+33.1 cm/s ). (indicate the direction with the sign of your answer.)
( v_{1f}= cm/s )
( v_{2f}= cm/s )

Explanation:

Step1: Use conservation of momentum and kinetic energy for elastic collision (equal mass)

For two objects of equal mass \(m_1 = m_2=m\) in an elastic collision, the velocity formulas are \(v_{1f}=v_{2i}\) and \(v_{2f}=v_{1i}\)

Given \(v_{1i}=+ 33.1\space cm/s\) and \(v_{2i}=-20.2\space cm/s\)

Step2: Calculate final velocities

\(v_{1f}=v_{2i}=-20.2\space cm/s\)

\(v_{2f}=v_{1i}=+33.1\space cm/s\)

Step3: For the exercise (different - mass case)

Let \(m_1 = 2m\) and \(m_2=m\)

Conservation of momentum: \(m_1v_{1i}+m_2v_{2i}=m_1v_{1f}+m_2v_{2f}\)

\(2m\times33.1+m\times(- 20.2)=2m\times v_{1f}+m\times v_{2f}\)

\(66.2m - 20.2m=2mv_{1f}+mv_{2f}\)

\(46m = 2mv_{1f}+mv_{2f}\)

\(46 = 2v_{1f}+v_{2f}\) (divide both sides by \(m\))

For elastic collision, \(v_{1i}-v_{2i}=v_{2f}-v_{1f}\)

\(33.1-(-20.2)=v_{2f}-v_{1f}\)

\(53.3=v_{2f}-v_{1f}\)

\(v_{2f}=v_{1f}+53.3\)

Substitute \(v_{2f}\) into \(46 = 2v_{1f}+v_{2f}\)

\(46=2v_{1f}+v_{1f}+53.3\)

\(3v_{1f}=46 - 53.3\)

\(3v_{1f}=-7.3\)

\(v_{1f}=-2.43\space cm/s\)

Then \(v_{2f}=v_{1f}+53.3=-2.43 + 53.3 = 50.87\space cm/s\)

Answer:

For the practice it:
\(v_{1f}=-20.2\space cm/s\)
\(v_{2f}=+33.1\space cm/s\)

For the exercise:
\(v_{1f}=-2.43\space cm/s\)
\(v_{2f}=+50.87\space cm/s\)