Question

Two wooden blocks of mass M\(_1\) and M\(_2\) rest on a frictionless table. A bullet of mass m is fired at M\(_1\) with speed v which embedded in it and the two together finally collide with M\(_2\). Find the velocity of M\(_2\) after collision. [Ignore any energy loss and treat the problem to be one dimensional]

• A. \( \frac{2mv}{M_1+M_2+m} \)
• B. \( \frac{mv}{M_1+M_2+m} \)
• C. \( \frac{(M_1+M_2+m)v}{M_1+M_2+m} \) (This simplifies to v)
• D. \( \frac{M_1+M_2}{M_1+M_2+m}v \)

Hint:

\textbf{Perfectly Inelastic Collision:} Objects stick together. Momentum is conserved. Kinetic energy is not conserved (some is lost).
\textbf{Perfectly Elastic Collision (1D):} Both momentum and kinetic energy are conserved. Velocities after collision: \(v_1 = \frac{(m_1-m_2)u_1 + 2m_2u_2}{m_1+m_2}\), \(v_2 = \frac{2m_1u_1 + (m_2-m_1)u_2}{m_1+m_2}\). If target \(m_2\) is initially at rest (\(u_2=0\)): \(v_2 = \frac{2m_1u_1}{m_1+m_2}\).
Interpret "embedding" as perfectly inelastic. "Ignore energy loss" is then likely applied to subsequent elastic interactions.

Answer 1

The Correct Option is A

The problem involves two stages of collision. "Ignore any energy loss" implies all collisions are perfectly elastic. However, "bullet ... embedded in it" means the first collision (bullet with M\(_1\)) is perfectly inelastic. This is a contradiction in the problem statement if "ignore any energy loss" applies to all stages. Let's assume "bullet embedded in it" defines the first collision as perfectly inelastic, and "ignore any energy loss" might refer to the second collision (M\(_1\)+m with M\(_2\)) being elastic, or it's a general (possibly flawed) instruction. Stage 1: Bullet (m) collides with M\(_1\) and embeds in it (perfectly inelastic collision). Initial momentum = \(mv + M_1(0) = mv\). Let \(V_1\) be the velocity of the combined mass \((M_1+m)\) after this collision. Final momentum = \((M_1+m)V_1\). By conservation of momentum: \(mv = (M_1+m)V_1 \Rightarrow V_1 = \frac{mv}{M_1+m}\). Stage 2: The combined mass \((M_1+m)\) with velocity \(V_1\) collides with M\(_2\) (initially at rest). The problem states "ignore any energy loss" and "treat the problem to be one dimensional". If this collision is perfectly elastic: Let \(m_A = M_1+m\) and \(m_B = M_2\). Initial velocities: \(u_A = V_1 = \frac{mv}{M_1+m}\), \(u_B = 0\). Final velocities: \(v_A, v_B\). We need to find \(v_B\) (velocity of M\(_2\) after collision). For a 1D elastic collision: \(v_B = \frac{2m_A u_A + (m_B-m_A)u_B}{m_A+m_B}\). Since \(u_B=0\): \(v_B = \frac{2m_A u_A}{m_A+m_B}\). Substitute \(m_A = M_1+m\) and \(u_A = \frac{mv}{M_1+m}\): \(v_B = \frac{2(M_1+m) \left(\frac{mv}{M_1+m}\right)}{(M_1+m) + M_2}\) \(v_B = \frac{2mv}{(M_1+m) + M_2} = \frac{2mv}{M_1+M_2+m}\). This matches option (a). This interpretation assumes the first collision is perfectly inelastic (embedding) and the second collision is perfectly elastic ("ignore any energy loss" applied to the second collision). This is the only way to arrive at option (a). If "ignore any energy loss" meant the entire process was somehow without loss despite embedding, the problem is physically inconsistent. The standard interpretation of "embedding" is a perfectly inelastic collision where kinetic energy is not conserved. What if the "two together finally collide with M\(_2\)" was also perfectly inelastic? Then \((M_1+m)V_1 + M_2(0) = (M_1+m+M_2)V_{final}\). \( (M_1+m) \frac{mv}{M_1+m} = (M_1+m+M_2)V_{final} \) \( mv = (M_1+M_2+m)V_{final} \) \( V_{final} = \frac{mv}{M_1+M_2+m} \). This is option (b). Option (b) would be the velocity of M\(_2\) (and M\(_1\)+m) if the second collision was also perfectly inelastic. The checkmark in the image is on option (a). This confirms the interpretation: 1. Bullet embeds in M\(_1\) (perfectly inelastic). 2. (M\(_1\)+m) collides elastically with M\(_2\). The phrase "ignore any energy loss" must specifically refer to the second collision. \[ \boxed{\frac{2mv}{M_1+M_2+m}} \]