Consider an elastic collision of a particle o | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) For an elastic collision between two particles of equal mass where one is initially at rest, we apply the conservation of linear momentum and kine

Vedclass · Accepted Answer

$90$

Vedclass · Answer

(B) For an elastic collision between two particles of equal mass where one is initially at rest, we apply the conservation of linear momentum and kinetic energy.Let the initial velocity be $\vec{u}_1 = u\hat{i}$ and $\vec{u}_2 = 0$.After collision, let the velocities be $\vec{v}_1$ and $\vec{v}_2$.Conservation of momentum: $m\vec{u}_1 = m\vec{v}_1 + m\vec{v}_2 \implies \vec{u}_1 = \vec{v}_1 + \vec{v}_2$.Squaring both sides: $u_1^2 = v_1^2 + v_2^2 + 2\vec{v}_1 \cdot \vec{v}_2$.Conservation of kinetic energy: $\frac{1}{2}mu_1^2 = \frac{1}{2}mv_1^2 + \frac{1}{2}mv_2^2 \implies u_1^2 = v_1^2 + v_2^2$.Comparing the two equations, we get $2\vec{v}_1 \cdot \vec{v}_2 = 0$.Since the velocities are non-zero, the dot product is zero, which implies $\vec{v}_1 \perp \vec{v}_2$.Thus, the angle between the two particles after the collision is $90^\circ$, meaning $	heta_1 + 	heta_2 = 90^\circ$.

Similar Questions

Explain the special cases of elastic collision in one dimension.

The coefficient of restitution for a perfectly elastic collision is

Collision takes place between two solid spheres denoted as $1$ and $2$. The initial velocities of the spheres are $u_1 = 3 \ m/s$ and $u_2 = 1.5 \ m/s$,and the final velocities are $v_1 = 2.5 \ m/s$ and $v_2 = 3.5 \ m/s$. The coefficient of restitution between the materials of the spheres is nearly:

$A$ proton of mass $m$ collides elastically with a particle of unknown mass $M$ at rest. After the collision, the proton and the unknown particle are seen moving at an angle of $90^\circ$ with respect to each other. The mass of the unknown particle is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module