A) 1.5 m/s
B) 2.0 m/s
C) 2.5 m/s
D) 3.0 m/s
Correct Answer: C
Solution :
Momentum of one piece \[=\frac{M}{4}\times 3\] Momentum of the other piece \[=\frac{M}{4}\times 4\] \ Resultant momentum \[=\sqrt{\frac{9{{M}^{2}}}{16}+{{M}^{2}}}=\frac{5M}{4}\] The third piece should also have the same momentum. Let its velocity be v, then \[\frac{5M}{4}=\frac{M}{2}\times v\] or \[v=\frac{5}{2}=2.5\,m/sec\]
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