A smooth sphere is moving on a horizontal sur | vedclass

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

Question

(C) The velocity of the sphere before impact is $\vec{v}_1 = 3 \hat{i} + \hat{j}$.The wall is parallel to the $\hat{j}$ axis,meaning the wall lies in

Vedclass · Accepted Answer

$-\hat{i} + \hat{j}$

Vedclass · Answer

(C) The velocity of the sphere before impact is $\vec{v}_1 = 3 \hat{i} + \hat{j}$.The wall is parallel to the $\hat{j}$ axis,meaning the wall lies in the $yz$-plane (normal to the $x$-axis).The component of velocity parallel to the wall ($\hat{j}$ component) remains unchanged during the collision because the impulse acts only along the normal to the wall.Thus,$v_{y, 	ext{after}} = v_{y, 	ext{before}} = 1 \hat{j}$.The component of velocity perpendicular to the wall ($\hat{i}$ component) changes according to the coefficient of restitution $e$.The velocity of approach along the normal is $u_x = 3$.The velocity of separation along the normal is $v_x = -e \cdot u_x = -\frac{1}{3} \cdot 3 = -1$.Therefore,the velocity vector after the collision is $\vec{v}_2 = -1 \hat{i} + 1 \hat{j} = -\hat{i} + \hat{j}$.

Similar Questions

$A$ body of mass $2 \,kg$ collides head-on with another body of mass $4 \,kg$. If the relative velocities of the bodies before and after collision are $10 \,ms^{-1}$ and $4 \,ms^{-1}$ respectively, the loss of kinetic energy of the system due to the collision is (in $\,J$)

$A$ $1.0 \ kg$ ball drops vertically onto a floor from a height of $25 \ cm$. It rebounds to a height of $4 \ cm$. The coefficient of restitution for the collision is

$A$ ball of mass $m$ is moving with a speed $V$ as shown in the figure. It undergoes an inelastic collision with a ball of mass $2m$ which was initially at rest. The velocity of the ball of mass $2m$ after the collision is given by

$A$ body of mass $m_1$ moving with a velocity $3 \, ms^{-1}$ collides with another body at rest of mass $m_2$. After the collision,the velocities of the two bodies are $2 \, ms^{-1}$ and $5 \, ms^{-1}$ respectively along the direction of motion of $m_1$. The ratio $\frac{m_1}{m_2}$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module