Imagine two billiard balls on a pool table. Ball A has a mass of 2 kilograms and ball
B has a mass of 3 kilograms. The initial velocity of ball A is 9 meters per second to
the right, and the initial velocity of the ball B is 6 meters per second to the left. The
final velocity of ball A is 9 meters per second to the left, while the final velocity of
ball B is 6 meters per second to the right.
1. Explain what happens to each ball after the collision. Why do you think this
occurs? Which of Newton’s laws does this represent?
2. What can you say about the total momentum before and after the collision?
3. What do you think would happen to the velocity of each ball after the
collision if the masses and initial velocities of each ball were the sam

1. The signs of the velocities of the balls were changed, this is a result of the elastic collision between the two balls. The law involved in here is the Law of Interaction.
2. During the elastic collision, the momentum is conserved.
3. The velocity would be lower. There is a tendency for these balls to stick together.

Answer Link

shirleywashington shirleywashington · Answer 2 · 2019-02-01T13:01:49+01:00

Explanation :

It is given that

mass of ball A [tex]m_A=2\ Kg[/tex]

mass of ball B, [tex]m_B=3\ Kg[/tex]

initial velocity of ball A, [tex]u_A=9\ m/s[/tex] (in right)

initial velocity of ball B, [tex]u_B=-6\ m/s[/tex] (in left)

final velocity of ball A, [tex]u_A=-9\ m/s[/tex] (in left)

final velocity of ball B, [tex]u_A=6\ m/s[/tex] (in right)

(1) After collision the velocities of balls gets exchanged.The velocity of ball A and B will become 9 m/s and 6 m/s but in opposite direction.

This represents Newton's third law.

(2) Initial momentum,[tex]p_i=m_Au_A+m_Bu_B[/tex]

[tex]P_i=2\times 9+3\times -6[/tex]

[tex]p_i=0[/tex]

Final momentum, [tex]p_f=m_Av_A+m_Bv_B[/tex]

[tex]p_f=2\times (-9)+3\times 6[/tex]

[tex]p_f=-18+18=0[/tex]

So, this shows momentum is conserved.

(3) If masses and initial velocities of each ball were same, then [tex]m_A=m_B=m[/tex] and [tex]u_A+u_B=u[/tex]

So, [tex]2mu=m(v_A+v_B)[/tex]

[tex]v_A=2u-v_B[/tex]

and [tex]v_B=2u-v_A[/tex]

This is required solution.