a body of mass 300g kept at rest breaks into two parts due to internal force. One part 200g found to move at speed act of 12 m/s towards north what will be the velocity of the other part and in which direction it will move.
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Answered by
153
The body is at rest and force is internal hence the body breaks due to equal force so,
when mass=200 g=0.2 kg
and velocity=12 m/s
Force=mass*velocity=0.2 kg*12 g=2.4 N
equal force on the other part so,
velocity=2.4 N/0.3 kg=8 m/s.
the other part will move in the opposite direction that is south.
when mass=200 g=0.2 kg
and velocity=12 m/s
Force=mass*velocity=0.2 kg*12 g=2.4 N
equal force on the other part so,
velocity=2.4 N/0.3 kg=8 m/s.
the other part will move in the opposite direction that is south.
Answered by
66
Initially the body was at rest .The linear momentum of the body is p =mu= 0 .The body breaks due to internal force. As the external force acting on it is zero , its linear momentum will remain constant that is zero .
P1=m1v1 = 200g×12m/s, towards east
the linear momentum of the Other path must be have the same magnitude in should be opposite in direction hit for moves towards the west if the speed is be to its linear momentum is
P2= m2V2 = g ×V2.
Now, m1v1=m2V2.
Thus, 200g×12m/s
= g×v2
or V2 =24m/s.
the velocity of the other part is 24 M per second towards the west .
P1=m1v1 = 200g×12m/s, towards east
the linear momentum of the Other path must be have the same magnitude in should be opposite in direction hit for moves towards the west if the speed is be to its linear momentum is
P2= m2V2 = g ×V2.
Now, m1v1=m2V2.
Thus, 200g×12m/s
= g×v2
or V2 =24m/s.
the velocity of the other part is 24 M per second towards the west .
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