A) 7.5 J
B) 5 J
C) 2.5 J
D) 10 J
Correct Answer: A
Solution :
This is a case of inelastic collision, in which momentum is conserved, and total energy. Let\[{{u}_{1}}\]be velocity of body of mass 5 kg, and\[{{u}_{2}}\]of 2.5 kg. Then Momentum before collision = Momentum after collision \[{{m}_{1}}{{u}_{1}}+{{m}_{2}}{{u}_{2}}=({{m}_{1}}+{{m}_{2}}){{v}_{1}}\] Given, \[{{m}_{1}}=5\,kg,\,{{m}_{2}}=2.5\,kg,\,{{u}_{2}}=0\] \[\therefore \] \[5{{u}_{1}}=(5+2.5)v\] \[\Rightarrow \] \[{{u}_{1}}=\frac{7.5}{5}v=1.5\,v\] ...(i) Also, kinetic energy after collision \[=\frac{1}{2}({{m}_{1}}+{{m}_{2}}){{v}^{2}}\] Given, \[5=\frac{1}{2}({{m}_{1}}+{{m}_{2}}){{v}^{2}}\] ... (ii) \[\Rightarrow \] \[10=7.5\text{ }{{v}^{2}}\] \[\Rightarrow \] \[{{v}^{2}}=\frac{10}{7.5}=\frac{4}{3}\] Kinetic energy of 1st body after collision \[=\frac{1}{2}{{m}_{1}}u_{1}^{2}\] \[=\frac{1}{2}\times 5\times {{(1.5v)}^{2}}\] \[=\frac{1}{2}\times 5\times {{(1.5)}^{2}}\times \frac{4}{3}=7.5\,J\] Note: In an inelastic collision, kinetic energy is not conserved.You need to login to perform this action.
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