Question

solution
write the conservation of momentum equation. because ( m_1 = m_2 ), we can cancel the masses, then substitute ( v_{1i}=+30.0 cm/s ) and ( v_{2i}=-20.0 cm/s ) (step 3).
next, apply conservation of energy in the linear equation (step 4).
now solve equations (1) and (3) simultaneously by adding them together (step 5).
substitute the answer for ( v_{2f} ) into equation (1).
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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.

Explanation:

• Momentum Conservation: The law of conservation of momentum states that the total momentum of an isolated system remains constant. If \(m_1 = m_2\) and \(v_{1f}=v_{2f} = 0\), then \(m_1v_{1i}+m_2v_{2i}=0\). But for an elastic collision (where kinetic energy is also conserved), the kinetic energy \(K=\frac{1}{2}m_1v_{1i}^2+\frac{1}{2}m_2v_{2i}^2\). If \(v_{1f} = v_{2f}=0\), then \(K_f = 0\), while \(K_i=\frac{1}{2}m_1v_{1i}^2+\frac{1}{2}m_2v_{2i}^2>0\) (unless \(v_{1i}=v_{2i} = 0\), but that's not a collision situation).
• Energy Conservation: In an elastic collision, kinetic energy is conserved. The kinetic energy of a moving object \(K=\frac{1}{2}mv^2\). If both final velocities are zero, the final kinetic energy \(K_f = 0\). But the initial kinetic energy \(K_i=\frac{1}{2}m_1v_{1i}^2+\frac{1}{2}m_2v_{2i}^2\) (for non - zero initial velocities) is non - zero. Also, from the conservation of momentum \(m_1v_{1i}+m_2v_{2i}=m_1v_{1f}+m_2v_{2f}\). If \(v_{1f} = v_{2f}=0\), then \(m_1v_{1i}+m_2v_{2i}=0\). But combining with kinetic energy conservation (for elastic collision \(v_{1i}-v_{2i}=-(v_{1f} - v_{2f})\)), we can show that non - zero initial velocities (in a non - trivial collision) cannot lead to both final velocities being zero.

Answer:

No, because energy is conserved and that requires one ball or both to be moving after the collision.