A ball is dropped from a height h_0. It under | vedclass

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

Question

(A) Let $v_0$ be the velocity of the ball just before the first collision and $u_1$ be the velocity just after the first collision. Then $u_1 = e v_0$

Vedclass · Accepted Answer

$e^n = \sqrt{\frac{h_n}{h_0}}$

Vedclass · Answer

(A) Let $v_0$ be the velocity of the ball just before the first collision and $u_1$ be the velocity just after the first collision. Then $u_1 = e v_0$.After the second collision,$u_2 = e u_1 = e^2 v_0$.Continuing this process,after $n$ collisions,the velocity $u_n$ is given by $u_n = e^n v_0$.We know that the velocity of a body falling from height $h$ is $v = \sqrt{2gh}$.Thus,$u_n = \sqrt{2gh_n}$ and $v_0 = \sqrt{2gh_0}$.Substituting these into the equation $u_n = e^n v_0$,we get $\sqrt{2gh_n} = e^n \sqrt{2gh_0}$.Dividing both sides by $\sqrt{2gh_0}$,we obtain $e^n = \sqrt{\frac{h_n}{h_0}}$.

$A$ ball is dropped from a height $h_0$. It undergoes $n$ collisions with the ground. If the velocity of the ball after $n$ rebounds is $u_n$ and it reaches a height $h_n$,which formula gives the coefficient of restitution $e$?

Similar Questions

$A$ $1 \,kg$ ball moving at $12 \,ms^{-1}$ collides with a $2 \,kg$ ball moving in the opposite direction at $24 \,ms^{-1}$. If the coefficient of restitution is $2/3$, then their velocities after the collision are

$A$ sphere of mass $m$ moving with velocity $u$ undergoes a collision with a stationary sphere of mass $m$. If the coefficient of restitution is $e = 1/2$,what is the ratio of the final velocities of the two spheres?

Particle $A$ moving with a velocity $v = 10 \ m/s$ experiences a head-on collision with a stationary particle $B$ of the same mass. As a result of the collision,the kinetic energy of the system decreases by $1 \%$. The speed of particle $A$ after the collision is: (in $m/s$)

$A$ rubber ball is released from a height of $5\, m$ above the floor. It bounces back repeatedly,always rising to $\frac{81}{100}$ of the height through which it falls. Find the average speed of the ball. (Take $g = 10\, m/s^2$)

$A$ ball of mass $m$,moving with a speed $2v_0$,collides inelastically $(e > 0)$ with an identical ball at rest. Show that
$(a)$ For head-on collision,both the balls move forward.
$(b)$ For a general collision,the angle between the two velocities of scattered balls is less than $90^o$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module