A) 1 : 4
B) 1 : 2
C) 1 : 1
D) 4 : 1
Correct Answer: B
Solution :
Key Idea: \[KE=\frac{1}{2}m{{v}^{2}},p=mv,\] Kinetic energy \[KE=\frac{1}{2}m{{v}^{2}}\] Linear momentum \[p=mv\] or \[KE=\frac{1}{2m}{{(mv)}^{2}}=\frac{{{p}^{2}}}{2m}\] or \[p=\sqrt{2mKE}\] or \[p\propto \sqrt{m}\] or \[\frac{{{p}_{1}}}{{{p}_{2}}}=\sqrt{\frac{{{m}_{1}}}{{{m}_{2}}}}\] \[\therefore \] \[\frac{{{p}_{1}}}{{{p}_{2}}}=\sqrt{\frac{m}{4m}}=\sqrt{\frac{1}{4}}=\frac{1}{2}\]
You need to login to perform this action.
You will be redirected in
3 sec
