A) 1.07 kJ
B) 2.14 Kj
C) 2.4 kJ
D) 4.8 kJ
Correct Answer: D
Solution :
From the law of conservation of momentum, when no external force acts upon a system of two (or more) bodies, then the total momentum of the system remains constant. Momentum before explosion = momentum after explosion Since bomb v at rest, its velocity is zero, hence \[\operatorname{mv}={{m}_{1}}{{\nu }_{1}}\,\,+\,\,{{m}_{2}}{{\nu }_{2}}\] \[3\,\,\times \,\,0\,\,=\,\,2{{\nu }_{1}}+1\,\,\times \,\,80\] \[\Rightarrow \,\,\,\,\,{{\nu }_{1}}=-\frac{80}{2}\,\,=\,\,-\,40\,m/s\] Total energy imparted is \[KE=\frac{1}{2}{{m}_{1}}\nu _{1}^{2}+\frac{1}{2}{{m}_{2}}\nu _{2}^{2}\] \[=\,\,\,\frac{1}{2} \times \,\,2\,\,\times \,\,{{\left( -\,40 \right)}^{2}}\,\,+\,\,\frac{1}{2}\,\,\times \,\,1\,\,\times \,\,{{\left( 80 \right)}^{2}}\] \[= 1600+3200= 4800 J\] \[=\text{ }4.8\text{ }kJ\] Note: Since velocity of second piece is negative, it indicates that the piece is moving in opposite direction to the direction of the other piece.You need to login to perform this action.
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