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Two objects of masses 1Kg and 3Kg have equal momentum. What is the ratio of their kinetic energies?
- a)3:1
- b)9:1
- c)1:1
- d)1:2
Correct answer is option 'A'. Can you explain this answer?
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Solution:
Given:
Mass of object 1, m1 = 1 kg
Mass of object 2, m2 = 3 kg
Momentum of both objects is equal.
Let the momentum be p.
p = m1v1 = m2v2
where v1 and v2 are the velocities of the two objects.
We know that kinetic energy is given by:
K.E. = (1/2)mv^2
where m is the mass of the object and v is its velocity.
Let the kinetic energies of the two objects be K1 and K2.
We can write:
K1 = (1/2)m1v1^2
K2 = (1/2)m2v2^2
Dividing K2 by K1, we get:
K2/K1 = (1/2)m2v2^2 / (1/2)m1v1^2
K2/K1 = m2v2^2 / m1v1^2
We know that p = m1v1 = m2v2
Therefore, v2/v1 = m1/m2
Substituting this in the above equation, we get:
K2/K1 = m2v2^2 / m1v1^2 = (m2/m1) * (v2/v1)^2 = (m2/m1) * (m1/m2)^2 = m2/m1
Therefore, K2/K1 = m2/m1 = 3/1
Hence, the ratio of their kinetic energies is 3:1, which is option A.
Given:
Mass of object 1, m1 = 1 kg
Mass of object 2, m2 = 3 kg
Momentum of both objects is equal.
Let the momentum be p.
p = m1v1 = m2v2
where v1 and v2 are the velocities of the two objects.
We know that kinetic energy is given by:
K.E. = (1/2)mv^2
where m is the mass of the object and v is its velocity.
Let the kinetic energies of the two objects be K1 and K2.
We can write:
K1 = (1/2)m1v1^2
K2 = (1/2)m2v2^2
Dividing K2 by K1, we get:
K2/K1 = (1/2)m2v2^2 / (1/2)m1v1^2
K2/K1 = m2v2^2 / m1v1^2
We know that p = m1v1 = m2v2
Therefore, v2/v1 = m1/m2
Substituting this in the above equation, we get:
K2/K1 = m2v2^2 / m1v1^2 = (m2/m1) * (v2/v1)^2 = (m2/m1) * (m1/m2)^2 = m2/m1
Therefore, K2/K1 = m2/m1 = 3/1
Hence, the ratio of their kinetic energies is 3:1, which is option A.
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