
A body of mass m is at rest. Another body of the same mass moving with velocity V makes head on elastic collision with the first body. After collision the first body starts to move with velocity
A. V
B. \[2V\]
C. Remains at rest
D. Not predictable
Answer
161.4k+ views
Hint: In order to solve this question, we will use the concept of elastic collision in which momentum of the system and kinetic energy of the system remains conserved before and after the collision, using this we will solve for the velocity of the first body after the collision.
Formula used:
Kinetic energy of a body is calculated as
$K.E = \dfrac{1}{2}m{v^2}$
And, momentum of a body is calculated as
$P = mv$
where m is the mass and v is the velocity of the body.
Complete step by step solution:
According to the question, we have given that initially first body of mass m was at rest and second body of same mass m collides with it with a velocity of v so initial momentum and kinetic energy for the system was,
\[{P_i} = mV \\
\Rightarrow K.{E_i} = \dfrac{1}{2}m{V^2} \\ \]
Now, after the collision the let first body moves with velocity v and second body moves with velocity v’ so final momentum and kinetic energy of the system is
\[{P_f} = mv + mv' \\
\Rightarrow K.{E_i} = \dfrac{1}{2}m{v^2} + \dfrac{1}{2}mv{'^2} \\ \]
Now, equating initial and final conditions of momentum and kinetic energy we get,
$mV = mv + mv' \\
\Rightarrow V = v + v' \to (i) \\ $
$\dfrac{1}{2}m{V^2} = \dfrac{1}{2}m{v^2} + \dfrac{1}{2}mv{'^2} \\
\Rightarrow {V^2} = {v^2} + v{'^2} \\ $
put the value of $v' = V - v$ from equation (i) in above equation and solving for v we get,
${V^2} = {v^2} + v{'^2} \\
\Rightarrow {V^2} - {v^2} = {V^2} + {v^2} - 2vV \\
\Rightarrow 2vV - 2{v^2} = 0 \\
\therefore v(V - v) = 0 $
Now, either $v = 0,v = V$ masses are the same, so energy must transfer from one body to another so the first body will start moving with velocity $v = V$ and the second body will come to rest due to head-on collision.
Hence, the correct answer is option A.
Note: It should be remembered that, it was head on collision and bodies were of same mass which leads to transfer of the whole energy of one body to another if they were of different mass then both bodies would start moving with different velocities after the collision.
Formula used:
Kinetic energy of a body is calculated as
$K.E = \dfrac{1}{2}m{v^2}$
And, momentum of a body is calculated as
$P = mv$
where m is the mass and v is the velocity of the body.
Complete step by step solution:
According to the question, we have given that initially first body of mass m was at rest and second body of same mass m collides with it with a velocity of v so initial momentum and kinetic energy for the system was,
\[{P_i} = mV \\
\Rightarrow K.{E_i} = \dfrac{1}{2}m{V^2} \\ \]
Now, after the collision the let first body moves with velocity v and second body moves with velocity v’ so final momentum and kinetic energy of the system is
\[{P_f} = mv + mv' \\
\Rightarrow K.{E_i} = \dfrac{1}{2}m{v^2} + \dfrac{1}{2}mv{'^2} \\ \]
Now, equating initial and final conditions of momentum and kinetic energy we get,
$mV = mv + mv' \\
\Rightarrow V = v + v' \to (i) \\ $
$\dfrac{1}{2}m{V^2} = \dfrac{1}{2}m{v^2} + \dfrac{1}{2}mv{'^2} \\
\Rightarrow {V^2} = {v^2} + v{'^2} \\ $
put the value of $v' = V - v$ from equation (i) in above equation and solving for v we get,
${V^2} = {v^2} + v{'^2} \\
\Rightarrow {V^2} - {v^2} = {V^2} + {v^2} - 2vV \\
\Rightarrow 2vV - 2{v^2} = 0 \\
\therefore v(V - v) = 0 $
Now, either $v = 0,v = V$ masses are the same, so energy must transfer from one body to another so the first body will start moving with velocity $v = V$ and the second body will come to rest due to head-on collision.
Hence, the correct answer is option A.
Note: It should be remembered that, it was head on collision and bodies were of same mass which leads to transfer of the whole energy of one body to another if they were of different mass then both bodies would start moving with different velocities after the collision.
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