A particle of mass m_1 collides with a partic | vedclass

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

Question

(A) Let the initial velocity of $m_1$ be $\vec{u}_1$ and final velocities be $\vec{v}_1$ and $\vec{v}_2$. Since the collision is elastic,kinetic energ

Vedclass · Accepted Answer

$1$

Vedclass · Answer

(A) Let the initial velocity of $m_1$ be $\vec{u}_1$ and final velocities be $\vec{v}_1$ and $\vec{v}_2$. Since the collision is elastic,kinetic energy and momentum are conserved.Conservation of momentum: $m_1 \vec{u}_1 = m_1 \vec{v}_1 + m_2 \vec{v}_2$.Squaring both sides: $m_1^2 u_1^2 = m_1^2 v_1^2 + m_2^2 v_2^2 + 2 m_1 m_2 \vec{v}_1 \cdot \vec{v}_2$.Since the particles move at $90^{\circ}$ to each other,$\vec{v}_1 \cdot \vec{v}_2 = 0$,so $m_1^2 u_1^2 = m_1^2 v_1^2 + m_2^2 v_2^2$.Conservation of kinetic energy: $\frac{1}{2} m_1 u_1^2 = \frac{1}{2} m_1 v_1^2 + \frac{1}{2} m_2 v_2^2$,which implies $m_1 u_1^2 = m_1 v_1^2 + m_2 v_2^2$.From energy conservation,$m_1(u_1^2 - v_1^2) = m_2 v_2^2$.From momentum conservation,$m_1^2(u_1^2 - v_1^2) = m_2^2 v_2^2$.Dividing the two equations: $m_1 = m_2$. Thus,the ratio $\frac{m_2}{m_1} = 1$.

$A$ particle of mass $m_1$ collides with a particle of mass $m_2$ at rest. After the elastic collision,the two particles move at an angle of $90^{\circ}$ with respect to each other. The ratio $\frac{m_2}{m_1}$ is

Similar Questions

$A$ particle of mass $1\, kg$ moving with velocity $1\, m/s$,collides elastically with another particle of mass $m$ initially at rest. In the collision,the particle of mass $1\, kg$ loses $\frac{3}{4}$ of its kinetic energy $(K.E.)$. The value of $m$ is:

Two equal masses $m_1$ and $m_2$ moving along the same straight line with velocities $+3 \, m/s$ and $-5 \, m/s$ respectively collide elastically. Their velocities after the collision will be respectively

After a head-on elastic collision between two balls of equal masses,one is observed to have a speed of $3 \, m/s$ along the positive $x$-axis and the other has a speed of $2 \, m/s$ along the negative $x$-axis. The original velocities of the balls are:

$A$ ball of mass $m$ moving with speed $u$ collides with a smooth horizontal surface at an angle $\theta$ with it as shown in the figure. The magnitude of the impulse imparted to the surface by the ball is [Coefficient of restitution of collision is $e$].

When a moving body collides with a stationary body of $n$ times its mass,then the amount of kinetic energy transferred to the stationary body is

Vedclass Test Series

Exam Paper Generator

Online Exam Module