170k views
4 votes
The collision between two gas molecules with different masses results in the velocity of the first molecule decreasing by 1/4. If the collision is elastic, which of the following statements is true?

A. The velocity of the second molecule will increase by a factor of 4.
B. The velocity of the second molecule will increase by a factor of 16.
C. The velocity of the second molecule will decrease by a factor of 16.
D. The velocity change of the second molecule depends on both the mass and the velocity of the first molecule.
Please explain why it is the right answer too

User Gwhn
by
6.2k points

1 Answer

4 votes

Answer: The correct option is D.

Step-by-step explanation: We are given two gas molecules with different masses. The collision between them is elastic, hence the Total Kinetic energy of the system is conserved.

Initial Kinetic Energy:


K.E_i=(1)/(2)m_1u_1^2+(1)/(2)m_2u_2^2

Final Kinetic Energy:


K.E_f=(1)/(2)m_1v_1^2+(1)/(2)m_2v_2^2

where,


m_1 = mass of first molecule


m_2 = mass of second molecule


u_1 = initial velocity of first molecule


u_2 = initial velocity of second molecule


v_1 = final velocity of first molecule


v_2 = final velocity of second molecule

Elastic Collision:


K.E_i=K.E_f

That is,


(1)/(2)m_1u_1^2+(1)/(2)m_2u_2^2=(1)/(2)m_1v_1^2+(1)/(2)m_2v_2^2


m_1u_1^2+m_2u_2^2=m_1v_1^2+m_2v_2^2 .....(1)

We are given in the question that the final velocity of first molecule is decreased by 1/4, which means:


v_1=(u_1)/(4)

Putting this value in equation 1 , we get


m_1u_1^2+m_2u_2^2=(m_1u_1^2)/(16)+m_2v_2^2


m_1u_1^2-(m_1u_1^2)/(16)+m_2u_2^2=m_2v_2^2


m_1((15u_1^2)/(16))+m_2u_2^2=m_2v_2^2

taking
m_2 on other side, we get


(m_1)/(m_2)((15u_1^2)/(16))+u_2^2=v_2^2

From the above relation, it is visible that the velocity change of the second molecule depends on both mass and velocity of the first molecule.

User Williamsi
by
5.6k points