Question

Two particles of masses m and 2m moving in opposite directions on a frictionless surface collide elastically with velocities v and 2v respectively. Find the fraction of kinetic energy lost by the colliding particles:

• A. Half
• B. One third
• C. zero
• D. One fourth

Answer 1

The correct option is C zero

Answer is C.
Let the final velocities of'm and 2 m be v1 and v2 respectively as, shown in the figure.
By conservation of momentum
$m\left(2v\right)+2m(-v)=m\left(v_1\right)+2m\left(v_2\right)$
or $0=m{v}_1+2m{v}_2$
or $v_1+2v_2=0$... (1)
and since the collision is elastic
$v_2-v_1=2v-(-v)$
or $v_2-v_1=3v$... (2)
Solving the above two equations, we get,
$v_2=v$ and $v_1=-2v$
i.e., the mass 2m returns with velocity v while the mass m returns with velocity 2 v in the direction shown in figure
The collision was elastic therefore, no kinetic energy is lost, KE loss = $K{E}_1-K{E}_1$
or $\left(\frac{1}{2}m(2v{)}^2+\frac{1}{2}\left(2m\right)(-v{)}^2\right)-\left(\frac{1}{2}m(2v{)}^2+\frac{1}{2}\left(2m\right)v^2\right)=0$.
Hence, the fraction of kinetic energy lost by the colliding particles is zero.