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Test Index
JEE Main 2013 Paper
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Section:
Physics
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© examsnet.com
Question : 86 of 90
Marks:
+1
,
-0
This question has Statement I and Statement II. Of the four choices given after the Statements, choose the one that best describes the two Statements.
Statement – I :
A point particle of mass m moving with speed v collides with stationary point particle of mass M. If the maximum energy loss possible is given as
f
(
1
2
m
v
2
)
f \left( \frac{1}{2} m v^{2} \right)
f
(
2
1
m
v
2
)
then
f
=
(
m
M
+
m
)
f = \left( \frac{m}{M+m} \right)
f
=
(
M
+
m
m
)
Statement – II:
Maximum energy loss occurs when the particles get stuck together as a result of thecollision.
[Main 2013]
Statement – I is true, Statement – II is true, Statement – II is not a correct explanation of Statement – I.
Statement – I is true, Statement – II is false.
Statement – I is false, Statement – II is true
Statement – I is true, Statement – II is true, Statement – II is a correct explanation of Statement – I
Validate
Solution:
Loss of energy is maximum when collision is inelastic as in an inelastic collision there will be maximum deformation.
KE in COM frame is
1
2
(
M
m
M
+
m
)
V
rel
2
\frac{1}{2} \left( \frac{Mm}{M+m} \right) V_{\text{rel}}^{2}
2
1
(
M
+
m
M
m
)
V
rel
2
KE
i
1
2
1
2
(
M
m
M
+
m
)
V
2
KE
f
=
0
(
∵
V
rel
=
0
)
\text{KE}_i \frac{1}{2} \frac{1}{2} \left( \frac{Mm}{M+m} \right) V^{2} \;\; \text{KE}_f = 0 \; (\because V_{\text{rel}} = 0)
KE
i
2
1
2
1
(
M
+
m
M
m
)
V
2
KE
f
=
0
(
∵
V
rel
=
0
)
Hence loss in energy is
1
2
(
M
m
M
+
m
)
V
2
\frac{1}{2} \left( \frac{Mm}{M+m} \right) V^{2}
2
1
(
M
+
m
M
m
)
V
2
⇒
f
=
M
M
+
m
\Rightarrow f = \frac{M}{M+m}
⇒
f
=
M
+
m
M
© examsnet.com
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