A) 3 m/s
B) 2 m/s
C) 1 m/s
D) \[\frac{2}{3}\,m/s\]
Correct Answer: D
Solution :
Key Idea: This is a case of inelastic collision. When two bodies stick together on colliding then only momentum is conserved (inelastic collision). Hence, momentum before collision = momentum after collision \[{{m}_{1}}\,{{v}_{1}}+{{m}_{2}}\,{{v}_{2}}=({{m}_{1}}+{{m}_{2}})v\] Given, \[{{m}_{1}}=2\,kg,\,{{v}_{1}}=3\,m/s,\,{{m}_{2}}=1\,kg\], \[{{v}_{2}}=-4\,m/s\] \[v=\frac{2\times 3-1\times 4}{2+1}=\frac{2}{3}\,m/s\].
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