
Particle A makes a head on elastic collision with another stationary particle B. They fly apart in opposite directions with equal speeds. The mass ratio will be-
A. \[\dfrac{1}{3}\]
B. \[\dfrac{1}{2}\]
C. \[\dfrac{1}{4}\]
D. \[\dfrac{2}{3}\]
Answer
165.9k+ views
Hint:According to the law of conservation of momentum, in an isolated system the total momentum of bodies acting upon each other does not change or remains constant unless an external force is applied to it. So, we can say that momentum can neither be created nor destroyed. Hence momentum before the collision is equal to the momentum after the collision. By using this we can find the relation between the bodies.
Formula used:
According to the conservation of momentum,
\[{m_1}{u_1} + {m_2}{u_2} = {m_1}{v_1} + {m_2}{v_2}\]
Where \[{u_1}\] and \[u_2\] are initial velocities and \[v_1\] and \[v_2\] are initial velocities.
Complete step by step solution:
For an elastic collision,
\[e = - \dfrac{{{v_1} - {v_2}}}{{{u_1} - {u_2}}} = 1\]
Let before collision conditions, the velocity of A be \[{u_1}\] and the velocity of B be \[{u_2} = 0\]. After collision conditions, the speed of A be \[{v_1}\] and the speed of B be\[{v_2}\].
According to the conservation of momentum,
\[{m_1}{u_1} + {m_2}{u_2} = {m_1}{v_1} + {m_2}{v_2}\]
Using \[{u_2} = 0\]
\[{m_1}{u_1} = {m_1}{v_1} - {m_2}{v_1}\] ……(1)
As we know that \[e = - \dfrac{{{v_1} - {v_2}}}{{{u_1} - {u_2}}} = 1\]
\[{v_2} - ( - {v_1}) = {u_1} - {u_2}\]
\[\Rightarrow 2{v_1} = {u_1}\] ……(2)
Putting equation 2 in 1, we get
\[2{m_1}{v_1} = {m_2}{v_1} - {m_1}{v_1}\]
\[\Rightarrow 3{m_1}{v_1} = {m_2}{v_1}\]
\[\therefore \dfrac{{{m_1}}}{{{m_2}}} = \dfrac{1}{3}\]
Therefore, the mass ratio will be \[\dfrac{{{m_1}}}{{{m_2}}} = \dfrac{1}{3}\].
Hence option A is the correct answer.
Note: When two bodies come in direct contact, then a collision must be occurring. Collision is the event in which the bodies exert forces on one another in about a relatively short time. An elastic collision is defined as one where there is no net loss of kinetic energy in the system due to the collision. A perfectly elastic collision is when two bodies collide but there is no loss in the total kinetic energy.
Formula used:
According to the conservation of momentum,
\[{m_1}{u_1} + {m_2}{u_2} = {m_1}{v_1} + {m_2}{v_2}\]
Where \[{u_1}\] and \[u_2\] are initial velocities and \[v_1\] and \[v_2\] are initial velocities.
Complete step by step solution:
For an elastic collision,
\[e = - \dfrac{{{v_1} - {v_2}}}{{{u_1} - {u_2}}} = 1\]
Let before collision conditions, the velocity of A be \[{u_1}\] and the velocity of B be \[{u_2} = 0\]. After collision conditions, the speed of A be \[{v_1}\] and the speed of B be\[{v_2}\].
According to the conservation of momentum,
\[{m_1}{u_1} + {m_2}{u_2} = {m_1}{v_1} + {m_2}{v_2}\]
Using \[{u_2} = 0\]
\[{m_1}{u_1} = {m_1}{v_1} - {m_2}{v_1}\] ……(1)
As we know that \[e = - \dfrac{{{v_1} - {v_2}}}{{{u_1} - {u_2}}} = 1\]
\[{v_2} - ( - {v_1}) = {u_1} - {u_2}\]
\[\Rightarrow 2{v_1} = {u_1}\] ……(2)
Putting equation 2 in 1, we get
\[2{m_1}{v_1} = {m_2}{v_1} - {m_1}{v_1}\]
\[\Rightarrow 3{m_1}{v_1} = {m_2}{v_1}\]
\[\therefore \dfrac{{{m_1}}}{{{m_2}}} = \dfrac{1}{3}\]
Therefore, the mass ratio will be \[\dfrac{{{m_1}}}{{{m_2}}} = \dfrac{1}{3}\].
Hence option A is the correct answer.
Note: When two bodies come in direct contact, then a collision must be occurring. Collision is the event in which the bodies exert forces on one another in about a relatively short time. An elastic collision is defined as one where there is no net loss of kinetic energy in the system due to the collision. A perfectly elastic collision is when two bodies collide but there is no loss in the total kinetic energy.
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