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GATE Mechanical Engineering (ME) 2020 Shift 2 Solved Paper
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© examsnet.com
Question : 46
Total: 65
A rigid block of mass
m
1
=
10
k
g
having velocity
v
0
=
2
m
/
s
strikes a stationary block of mass
m
2
=
30
k
g
after travelling
1
m
along a frictionless horizontal surface as shown in the figure.
The two masses stick together and jointly move by a distance of
0.25
m
further along the same frictionless surface, before they touch the mass-less buffer that is connected to the rigid vertical wall by means of a linear spring having a spring constant
k
=
10
5
N
/
m
. The maximum deflection of the spring is
c
m
(round off to 2 decimal places).
Your Answer:
Validate
Solution:
m
1
=
10
k
g
v
0
=
2
m
/
s
m
2
=
30
k
g
Since given surfaces are friction less surface. Thus, both energy and linear momentum will be conserve.
k
=
10
5
N
/
m
After collision both mass sticks together. Thus,
10
×
2
+
30
×
0
=
(
10
+
30
)
V
common
20
=
40
V
common
V
common
=
1
2
m
/
s
1
2
(
m
1
+
m
2
)
V
common
2
=
1
2
k
x
2
40
×
(
1
2
)
2
=
10
5
×
x
2
10
×
10
−
5
=
x
2
x
=
10
−
2
m
x
=
1
c
m
© examsnet.com
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