A) \[\sqrt{2}:2\]
B) 1 : 4
C) 1 : 2
D) 2 : 1
Correct Answer: C
Solution :
Linear momentum \[p=\sqrt{2m{{E}_{k}}}\] or \[p\propto \sqrt{m}\] (As K.E. = constant) Hence \[\frac{{{p}_{p}}}{{{p}_{\alpha }}}=\sqrt{\frac{{{m}_{p}}}{{{m}_{\alpha }}}}=\sqrt{\frac{1}{4}}=\frac{1}{2}\] so \[{{p}_{p}}:{{p}_{\alpha }}=1:2\]
You need to login to perform this action.
You will be redirected in
3 sec
