A block A of mass M moving with speed u colli | vedclass

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(C) $1$. In the first elastic collision between block $A$ and block $B$:$V_A = \left(\frac{M-m}{M+m}\right)u$$V_B = \left(\frac{2M}{M+m}\right)u$$2$.

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$\left(\frac{m}{M+m}\right)u$

Vedclass · Answer

(C) $1$. In the first elastic collision between block $A$ and block $B$:$V_A = \left(\frac{M-m}{M+m}\right)u$$V_B = \left(\frac{2M}{M+m}\right)u$$2$. After the collision,block $B$ moves towards block $C$ and compresses the spring. At the time of maximum compression,the velocities of blocks $B$ and $C$ become equal (let this velocity be $v$).$3$. By the law of conservation of linear momentum for the system of blocks $B$ and $C$:$m V_B = (m + m)v$$m \left(\frac{2M}{M+m}\right)u = 2mv$$v = \left(\frac{M}{M+m}\right)u$$4$. The velocity of block $C$ with respect to block $A$ is:$V_{CA} = V_C - V_A = v - V_A$$V_{CA} = \left(\frac{M}{M+m}\right)u - \left(\frac{M-m}{M+m}\right)u$$V_{CA} = \left(\frac{M - M + m}{M+m}\right)u = \left(\frac{m}{M+m}\right)u$

$A$ block $A$ of mass $M$ moving with speed $u$ collides elastically with block $B$ of mass $m$ which is connected to block $C$ of mass $m$ with a spring. When the compression in the spring is maximum,the velocity of block $C$ with respect to block $A$ is (neglect friction):

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