
An object of mass m1 collides with another object of mass m2, which is at rest. After the collision, the objects move at equal speeds in opposite directions. The ratio of the masses m2: m1 is:
a. 2:1
b. 1:1
c. 1:2
d. 3:1
Answer
162k+ views
Hint: Here conservation of linear momentum is obeyed and with the help of coefficient of restitution we can find the ratio ${m_1}:{_m2}$. For elastic collision velocity of separation (after collision) is equal to velocity of approach (before collision), the ratio of which is called coefficient of restitution.
Formula used:
From the conservation of linear momentum,
Total linear momentum before collision= Total linear momentum after collision
\[{{m}_{1}}{{u}_{1}}+{{m}_{2}}{{u}_{2}}={{m}_{1}}{{v}_{1}}+{{m}_{2}}{{v}_{2}}\]
Complete answer:
Let ${u_1}$ and ${u_2}$ be the initial velocity of an object of mass ${m_1}$ and mass ${m_2}$ respectively. Let ‘v’ be the final velocity of the object of mass m1 and ‘-v’ be the velocity of object having mass $m_2$ respectively.
From the conservation of linear momentum,
Total linear momentum before collision= Total linear momentum after collision
\[{{m}_{1}}{{u}_{1}}+{{m}_{2}}{{u}_{2}}={{m}_{1}}{{v}_{1}}+{{m}_{2}}{{v}_{2}}\]
From the question, it is given that
${{u}_{2}}=0$
On substituting we get,
${{m}_{1}}:{{m}_{2}}=3:1$ ${{m}_{1}}{{u}_{1}}=v\left( {{m}_{2}}-{{m}_{1}} \right)$
We have coefficient of restitution,
$e=\frac{v-(-v)}{{{u}_{1}}}=\frac{2v}{{{u}_{1}}}$
Given e=1,
$\therefore {{u}_{1}}=2v$
${{m}_{1}}2v=v\left( {{m}_{2}}-{{m}_{1}} \right)$
$2{{m}_{1}}={{m}_{2}}-{{m}_{1}}$
On further solving we get the ratio of masses as
${{m}_{1}}:{{m}_{2}}=3:1$
Therefore, the answer is: (d).
Note: Coefficient of restitution is simply the ratio of final relative velocity to that of initial. Value of the coefficient of restitution lies between 0 and 1. For perfectly elastic collision, the value of coefficient of restitution is 1. For inelastic collision, the value of coefficient of restitution is 0. This problem can also be solved with the help of conservation of kinetic energy. That is kinetic energy before collision is equal to kinetic energy after collision and also linear momentum will be already conserved. On solving the equation, we get the value of relation between initial and final velocities and on substituting we get the ratio of masses.
Formula used:
From the conservation of linear momentum,
Total linear momentum before collision= Total linear momentum after collision
\[{{m}_{1}}{{u}_{1}}+{{m}_{2}}{{u}_{2}}={{m}_{1}}{{v}_{1}}+{{m}_{2}}{{v}_{2}}\]
Complete answer:
Let ${u_1}$ and ${u_2}$ be the initial velocity of an object of mass ${m_1}$ and mass ${m_2}$ respectively. Let ‘v’ be the final velocity of the object of mass m1 and ‘-v’ be the velocity of object having mass $m_2$ respectively.
From the conservation of linear momentum,
Total linear momentum before collision= Total linear momentum after collision
\[{{m}_{1}}{{u}_{1}}+{{m}_{2}}{{u}_{2}}={{m}_{1}}{{v}_{1}}+{{m}_{2}}{{v}_{2}}\]
From the question, it is given that
${{u}_{2}}=0$
On substituting we get,
${{m}_{1}}:{{m}_{2}}=3:1$ ${{m}_{1}}{{u}_{1}}=v\left( {{m}_{2}}-{{m}_{1}} \right)$
We have coefficient of restitution,
$e=\frac{v-(-v)}{{{u}_{1}}}=\frac{2v}{{{u}_{1}}}$
Given e=1,
$\therefore {{u}_{1}}=2v$
${{m}_{1}}2v=v\left( {{m}_{2}}-{{m}_{1}} \right)$
$2{{m}_{1}}={{m}_{2}}-{{m}_{1}}$
On further solving we get the ratio of masses as
${{m}_{1}}:{{m}_{2}}=3:1$
Therefore, the answer is: (d).
Note: Coefficient of restitution is simply the ratio of final relative velocity to that of initial. Value of the coefficient of restitution lies between 0 and 1. For perfectly elastic collision, the value of coefficient of restitution is 1. For inelastic collision, the value of coefficient of restitution is 0. This problem can also be solved with the help of conservation of kinetic energy. That is kinetic energy before collision is equal to kinetic energy after collision and also linear momentum will be already conserved. On solving the equation, we get the value of relation between initial and final velocities and on substituting we get the ratio of masses.
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