
A body of mass 300 gram at rest breaks into two parts due to internal forces. One part of mass 200 gram is found to move at a velocity of 12 m/s towards east. Find the velocity of the other part.
(A) 24 m/s towards west
(B) 14 m/s towards east
(C) 24 m/s towards north
(D) 54 m/s towards south
Answer
549.6k+ views
Hint: The principle of conservation of linear momentum and Newton’s third law are consistent with each other.
$ \left[ {{m_1}{v_1} + {m_2}{v_2} = 0} \right] $
It means that net external force will be zero after application of law of conservation of linear momentum.
Complete step by step answer:
Mass of first part, $ {m_1} = 200g $
Velocity of first part towards east, $ {v_1} = 12\dfrac{m}{s} $
Hence, mass of second part = Total mass of body - 1st body’s mass
$ {m_2} = 300 - 200 $
$ {m_2} = 100g $
And its velocity, v =?
According law of conservation of linear motion,
$ \left[ {{m_1}{v_1} + {m_2}{v_2} = 0} \right] $
Mass of first body x velocity of first body + mass of second body x velocity of second body = 0
$ \left[ {200 \times 12 + 100 \times {v_2} = 0} \right] $
$ \left[ {2400 + 100{v_2} = 0} \right] $
$ {v_2} = - \dfrac{{2400}}{{100}} $
$ \left[ {{v_2} = - 24\dfrac{m}{s}} \right] $
Hence, the second part has gone in the west direction with a velocity of 24 m/s.
Hence, the correct option is A (24 m/s towards west).
This is a common application of the principle of conservation of linear momentum, that is: A person is staying on a frictionless surface at rest, his momentum will be zero. As well as if he throws an object, he goes in his opposite direction. Hence, the total momentum will be zero of a person or thrown an object due to the opposite direction.
Note:
Basically, Momentum is the product of Mass and velocity. That is, $ \left[ {P = mv} \right] $
The total sum of linear momentum of the body remains constant and is not influenced by their action and reaction. In the case of two bodies, total momentum of the body is equal before influence or after influence. The law of linear momentum is the universal law that is applied on both type systems, that are – Microscopic system and Macroscopic system.
$ \left[ {{m_1}{v_1} + {m_2}{v_2} = 0} \right] $
It means that net external force will be zero after application of law of conservation of linear momentum.
Complete step by step answer:
Mass of first part, $ {m_1} = 200g $
Velocity of first part towards east, $ {v_1} = 12\dfrac{m}{s} $
Hence, mass of second part = Total mass of body - 1st body’s mass
$ {m_2} = 300 - 200 $
$ {m_2} = 100g $
And its velocity, v =?
According law of conservation of linear motion,
$ \left[ {{m_1}{v_1} + {m_2}{v_2} = 0} \right] $
Mass of first body x velocity of first body + mass of second body x velocity of second body = 0
$ \left[ {200 \times 12 + 100 \times {v_2} = 0} \right] $
$ \left[ {2400 + 100{v_2} = 0} \right] $
$ {v_2} = - \dfrac{{2400}}{{100}} $
$ \left[ {{v_2} = - 24\dfrac{m}{s}} \right] $
Hence, the second part has gone in the west direction with a velocity of 24 m/s.
Hence, the correct option is A (24 m/s towards west).
This is a common application of the principle of conservation of linear momentum, that is: A person is staying on a frictionless surface at rest, his momentum will be zero. As well as if he throws an object, he goes in his opposite direction. Hence, the total momentum will be zero of a person or thrown an object due to the opposite direction.
Note:
Basically, Momentum is the product of Mass and velocity. That is, $ \left[ {P = mv} \right] $
The total sum of linear momentum of the body remains constant and is not influenced by their action and reaction. In the case of two bodies, total momentum of the body is equal before influence or after influence. The law of linear momentum is the universal law that is applied on both type systems, that are – Microscopic system and Macroscopic system.
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