JEE Main 12 Jan 2019 Evening Question 3
Question: An alpha-particle of mass m suffers -dimensional elastic collision with a nucleus at rest of unknown mass. It is scattered directly backwards losing, 64% of its initial kinetic energy. The mass of the nucleus is
[JEE Main Online Paper Held On 12-Jan-2019 Evening]
Options:
A) 4 m
B) 1.5 m
C) 3.5 m
D) 2 m
Show Answer
Answer:
Correct Answer: A
Solution:
-
Let M be the mass of the nucleus. Applying conservation of linear momentum, $ mv=mv _1+Mv _2 $ …(i)
Also, $ \frac{1}{2}mv_1^{2}=\frac{36}{100}\frac{1}{2}mv^{2}\Rightarrow v _1=\frac{6}{10}v $ …(i)Applying conservation of kinetic energy, $ \frac{1}{2}mv^{2}=\frac{1}{2}mv_1^{2}+\frac{1}{2}Mv_2^{2} $
$ \Rightarrow \frac{1}{2}Mv_2^{2}=\frac{64}{100}\frac{1}{2}mv^{2}\Rightarrow v _2=\frac{8}{10}v\sqrt{\frac{m}{M}} $
Substituting (ii) and (iii) in eqn. (i) $ mv=-( \frac{6}{10}v )m+M( \frac{8}{10}v\sqrt{\frac{m}{M}} )\Rightarrow \frac{16}{10}mv=\frac{8}{10}v\sqrt{mM} $
$ \Rightarrow M=4m $
