Question
A mass m moving with a speed u collides with a similar mass m at rest, elastically and obliquely. Prove that they will move in directions making an angle $\frac{\pi}{2}$ with each other.
✓
Answer
Let the equal masses be m. Let the velocity at their motion initially be u and zero. Since momentum is a vector, we have, $\text{mu}=\text{mv}_1\cos\theta_1+\text{mv}_2\cos\theta_2$ i.e., $\text{u}=\text{v}_1\cos\theta_1+\text{v}_2\cos\theta_2\cdots\text{(i)}$ $0=\text{v}_1\sin\theta_1+\text{v}_2\sin\theta_2\cdots\text{(ii)}$ based on the conservation of momentum.
Being an elastic collision, kinetic energy is also conserved
$\therefore\frac{1}{2}\text{mu}^2=\frac{1}{2}\text{mv}_1^2+\frac{1}{2}\text{mv}_2^2$
$\therefore\text{u}^2=\text{v}_1^2+\text{v}_2^2\cdots\text{(iii)}$
Squaring and adding (i) and (ii) we have,
$\text{u}^2=\text{v}_1^2+\text{v}_2^2+2\text{v}_1\text{v}_2$
$(\cos\theta_1\cos\theta_2-\sin\theta_1\sin\theta_2)$
$\text{u}^2=\text{v}_1^2+\text{v}_2^2+2\text{v}_1\text{v}_2\cos(\theta_1+\theta_2)$
Using (iii) in this equation, we have
$2\text{v}_1\text{v}_2\cos(\theta_1+\theta_2)=0$
$\therefore\cos(\theta_1+\theta_2)=0$
$\theta_1+\theta_2=\frac{\pi}{2}$
i.e., the masses move at right angles after the collision.
Being an elastic collision, kinetic energy is also conserved
$\therefore\frac{1}{2}\text{mu}^2=\frac{1}{2}\text{mv}_1^2+\frac{1}{2}\text{mv}_2^2$
$\therefore\text{u}^2=\text{v}_1^2+\text{v}_2^2\cdots\text{(iii)}$
Squaring and adding (i) and (ii) we have,
$\text{u}^2=\text{v}_1^2+\text{v}_2^2+2\text{v}_1\text{v}_2$
$(\cos\theta_1\cos\theta_2-\sin\theta_1\sin\theta_2)$
$\text{u}^2=\text{v}_1^2+\text{v}_2^2+2\text{v}_1\text{v}_2\cos(\theta_1+\theta_2)$
Using (iii) in this equation, we have
$2\text{v}_1\text{v}_2\cos(\theta_1+\theta_2)=0$
$\therefore\cos(\theta_1+\theta_2)=0$
$\theta_1+\theta_2=\frac{\pi}{2}$
i.e., the masses move at right angles after the collision.
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