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JEE Main Physics Class 11 Work, Energy and Power Part 1 Questions
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© examsnet.com
Question : 38
Total: 100
An alpha-particle of mass
m
suffers 1 -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 :
[12 Jan. 2019 II]
2m
3.5m
1.5m
4m
Validate
Solution:
Using conservation of momentum,
m
v
0
=
m
v
2
−
m
v
1
1
2
m
v
1
2
=
0.36
×
1
2
m
v
0
2
⇒
v
1
=
0.6
v
0
The collision is elastic. So,
1
2
M
V
2
2
=
0.64
×
1
2
m
v
0
2
[
∴
M
=
mass of nucleus
]
⇒
V
2
=
√
m
M
×
0.8
V
0
m
V
0
=
√
m
M
×
0.8
V
0
−
m
×
0.6
V
0
⇒
1.6
m
=
0.8
√
m
M
⇒
4
m
2
=
m
M
∴
M
=
4
m
© examsnet.com
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