A) 900 J
B) 300 J
C) 600 J
D) 1200 J
Correct Answer: C
Solution :
If v' is final velocity of wagon, then applying principle of conservation of linear momentum, we get, \[5\times {{10}^{3}}\times 1.2=(5\times {{10}^{3}}+{{10}^{3}})\times v'\] \[v'=1\,m{{s}^{-1}}\] Change in KE \[=\frac{1}{2}(6\times {{10}^{3}})\times {{1}^{2}}-\frac{1}{2}(5\times {{10}^{3}}){{(1.2)}^{2}}\] \[=600\,J\]You need to login to perform this action.
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