Question

c. break at that point.
d. not extend again.

1. two masses, a and b have the same momentum. mass b can have more kinetic energy than a if it

a. is moving faster than a.
b. has the same mass as a.
c. has less mass than a.
d. is moving at the same speed as a.

Explanation:

1. Recall the formulas for momentum (\(p = mv\)) and kinetic energy (\(K=\frac{1}{2}mv^{2}\)). Given \(p_A = p_B\), so \(m_Av_A=m_Bv_B\).
2. Express kinetic energy in terms of momentum: \(K = \frac{p^{2}}{2m}\) (derived from \(p = mv\Rightarrow v=\frac{p}{m}\), substitute into \(K=\frac{1}{2}mv^{2}\) to get \(K=\frac{p^{2}}{2m}\)).
3. Since \(p_A = p_B\) (let \(p = p_A = p_B\)), \(K\) is inversely proportional to \(m\) (\(K\propto\frac{1}{m}\) when \(p\) is constant). So for \(K_B>K_A\), \(m_B\) must be less than \(m_A\) (because smaller mass gives larger kinetic energy when momentum is equal).

• Option A: If \(v_B > v_A\) with \(p_A = p_B\), from \(p = mv\), \(m_B\) would be less than \(m_A\), but this is a consequence of \(m_B K_A\). The key relation is from \(K=\frac{p^{2}}{2m}\).
• Option B: If \(m_B=m_A\), then from \(p = mv\), \(v_B = v_A\) and \(K_B = K_A\).
• Option D: If \(v_B = v_A\), from \(p = mv\), \(m_B = m_A\) and \(K_B = K_A\).

Answer:

C. has less mass than A.