Physics - Center of Mass Question with Solution | TestHub

Solution:

Let $m_1=m$ and $m_2=m$ be the masses of the two particles.

Initial velocity of the first particle is $u_1=u$.

Initial velocity of the second particle is $u_2=0$.

After collision, let the velocities be $v_1$ and $v_2$.

The angles made with the initial direction are ${\theta }_1$ and ${\theta }_2$.

For an elastic collision, both momentum and kinetic energy are conserved.

Conservation of momentum in the x-direction (initial direction of motion):

$m{u}_1+m{u}_2=m{v}_1\cos {\theta }_1+m{v}_2\cos {\theta }_2$

$mu+m(0)=m{v}_1\cos {\theta }_1+m{v}_2\cos {\theta }_2$

$$u=v_1\cos {\theta }_1+v_2\cos {\theta }_2(1)$$

Conservation of momentum in the y-direction (perpendicular to initial direction):

$m{u}_1\sin 0+m{u}_2\sin 0=m{v}_1\sin {\theta }_1-m{v}_2\sin {\theta }_2$ (assuming ${\theta }_2$ is below the x-axis)

$$0=v_1\sin {\theta }_1-v_2\sin {\theta }_2(2)$$

From (2), $v_1\sin {\theta }_1=v_2\sin {\theta }_2$.

Conservation of kinetic energy:

$\frac{1}{2}m{u}_1^2+\frac{1}{2}m{u}_2^2=\frac{1}{2}m{v}_1^2+\frac{1}{2}m{v}_2^2$

$\frac{1}{2}m{u}^2+\frac{1}{2}m(0{)}^2=\frac{1}{2}m{v}_1^2+\frac{1}{2}m{v}_2^2$

$$u^2=v_1^2+v_2^2(3)$$

Solving the equations ${\theta }_1+{\theta }_2$=90