A) \[\frac{3}{4}m/s\]
B) \[\frac{1}{3}m/s\]
C) \[\frac{3}{2}m/s\]
D) \[\frac{2}{3}m/s\]
Correct Answer: C
Solution :
From law of conservation of momentum Momentum before collision = momentum after collision \[\therefore \] \[{{M}_{1}}{{u}_{1}}+{{M}_{2}}{{u}_{2}}=({{M}_{1}}+{{M}_{2}})v\] Given, \[{{M}_{1}}=10\,kg\], \[{{u}_{1}}=2\,m/s\], \[{{M}_{2}}=20\,kg,\,{{u}_{2}}=0\] \[\therefore \] \[10\times 2=(10+20)v\] \[\Rightarrow \] \[v=\frac{20}{30}=\frac{2}{3}m/s\]
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