Solve the following :

Question:

The blocks shown in figure have equal masses. The surface of $A$ is smooth but that of $B$ has a friction coefficient of $0.10$ with the floor. Block $A$ is moving at a speed of $10 \mathrm{~m} / \mathrm{s}$ towards $B$ which is kept at rest. Find the distance travelled by $B$ if (a) the collision is perfectly elastic and (b) the collision is perfectly inelastic. Take $g=10^{m} / \mathrm{s}^{2}$.

Solution:

(a) Use C.O.L.M

$m_{A}(10)=m_{A} V_{1}+m_{B} V_{2} \quad\left\{m_{A}=m_{B}\right\}$

$V_{1}+V_{2}=10-(1)$

Use C.O.E.L,

Solving (1) and (2)

$10=\sqrt{100-V_{2}^{2}}+V_{2}$

$2 V_{2}^{2}-20 V_{2}=0 \Rightarrow V_{2}=(0,10) V_{1}=(10,0)$

If $V_{1}=0, V_{2}=10$

$m_{B} a_{B}=m_{B} \mu g=0.1 \times 10=1 \mathrm{~m} / \mathrm{s}^{2}$

Use kinematics law,

$v^{2}=u^{2}+2 a x$

$x=\frac{(10)^{2}}{2 \times 1}=50$

(b) If it is inelastic,

C.O.L.M $\Rightarrow m_{A}(10)=\left(m_{A}+m_{B}\right) V$

$m_{A}(10)=2 m V$

.$V=\frac{10}{2}=5 \mathrm{~m} / \mathrm{s}$

$2 m a_{B}=m \mu g \Rightarrow a_{B}=\frac{10}{2 \times 10}=0.5 \mathrm{~m} / \mathrm{s}^{2}$

$x=\frac{V^{2}}{2 a}=\frac{25}{2 \times \frac{1}{2}}=25$

 

 

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