Question

a group of particles has a net momentum equal to zero. which of the following must be true?
i. the speed of each particle is zero.
ii. the velocity of each particle is zero.
iii. the mass of each particle must be equal.
iv. the kinetic energy of each particle must be zero.
o i and iv
o iii only
o i only
o i and iii
o ii only
o iv only
o none of the above is true.

Explanation:

• For I: If particles move in opposite directions with appropriate speeds (e.g., two particles of masses \(m_1 = m_2=m\), \(v_1 = v\) and \(v_2=-v\)), net momentum \(p = m_1v_1 + m_2v_2=mv - mv = 0\), but individual speeds (\(\vert v_1\vert=\vert v_2\vert = v

eq0\)) are non - zero.

• For II: Similar to I, velocity is a vector. Net momentum \(\vec{P}=\sum_{i = 1}^{n}m_i\vec{v}_i = 0\) does not imply \(\vec{v}_i=0\) for all \(i\). For example, two particles with \(m_1\vec{v}_1=-m_2\vec{v}_2\) (non - zero vectors) give \(\vec{P} = 0\).
• For III: Consider two particles \(m_1 = 2m\), \(v_1=v\) and \(m_2=m\), \(v_2 = - 2v\). Then \(p=m_1v_1+m_2v_2=2mv-2mv = 0\), but \(m_1

eq m_2\).

• For IV: Kinetic energy \(K_i=\frac{1}{2}m_iv_i^{2}\). From the example in I (non - zero \(v_i\)), \(K_i=\frac{1}{2}mv^{2}

eq0\) for each particle.

Answer:

None of the above is true.