Two magnets are placed at points A and B on a frictionless horizontal table. Their masses are 10g and 6g respectively. The magnets are released and they move on the table due to mutual attraction and collide. When the first magnet reaches a particular point ,its speed is found to be 3m/s. Find the speed of the other magnet at that instant.
Answers
Answer:
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Explanation:
Given - m
1
/m
2
=1:2 or m
1
=m,m
2
=2m
l
1
/l
2
=1:2 or l
1
=l,l
2
=2l
Moment of inertia of first bar magnet ,
I
1
=m
1
l
1
2
/12=ml
2
/12 ........................eq1
moment of inertia of second bar magnet ,
I
2
=m
2
l
2
2
/12=2m(2l)
2
/12=8ml
2
/12 .......................eq2
Maximum torque on a bar magnet in a magnetic field B is given by
τ=MB
As given , τ
1
=τ
2
or M
1
B
1
=M
2
B
2
.....................eq3
Now , the time period of oscillation of a bar magnet is given by ,
T=2π
I/MB
Therefore time period of oscillation of first bar magnet is ,
T
1
=2π
I
1
/M
1
B
1
and time period of oscillation of second bar magnet is ,
T
2
=2π
I
2
/M
2
B
2
=2π
I
2
/M
1
B
1
(Bt eq3)
dividing T
1
by T
2
T
1
/T
2
=
I
1
/I
2
=
8ml
2
/12
ml
2
/12
=1/2
2
Answer:
1 metre /s
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