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JEE Main 08 Apr 2018 Code A Solved Paper
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© examsnet.com
Question : 5
Total: 90
In a collinear collision, a particle with an initial speed
v
o
strikes a stationary particle of the same mass. If the final total kinetic energy is 50% greater than the original kinetic energy, the magnitude of the relative velocity between the two particles, after collision, is:
v
o
4
√
2
v
o
v
o
2
v
o
√
2
Validate
Solution:
Momentum conservation
m
v
0
=
m
(
v
1
−
v
2
)
v
1
=
v
2
=
v
0
.............(1)
1
2
m
(
v
1
2
+
v
2
2
)
−
1
2
m
v
0
2
=
1
2
×
1
2
m
v
0
2
v
1
2
+
v
2
2
=
3
2
v
0
2
.............(2)
(
v
1
−
v
2
)
2
=
v
0
2
v
1
2
+
v
2
2
−
2
v
1
v
2
=
v
0
2
.........(3)
(2) - (3)
⇒
2
v
1
v
2
=
v
0
2
2
⇒
4
v
1
v
2
=
v
0
2
(
v
1
+
v
2
)
=
√
(
v
1
−
v
2
)
2
+
4
v
1
v
2
Relative velocity after collision
=
√
v
0
2
+
v
0
2
=
√
2
v
0
© examsnet.com
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