A particle falls from a height h upon a fixed | vedclass

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

Question

(A) When a particle falls from height $h$,it hits the ground with velocity $v = \sqrt{2gh}$.After the first collision,it rebounds to a height $h_1 = e

Vedclass · Accepted Answer

$h\left( {\frac{{1 + {e^2}}}{{1 - {e^2}}}} \right)$

Vedclass · Answer

(A) When a particle falls from height $h$,it hits the ground with velocity $v = \sqrt{2gh}$.After the first collision,it rebounds to a height $h_1 = e^2h$.After the second collision,it rebounds to a height $h_2 = e^2h_1 = e^4h$.In general,the height after the $n^{th}$ rebound is $h_n = e^{2n}h$.The total distance $D$ travelled by the particle is the sum of the initial downward distance and twice the sum of all subsequent rebound heights (since it goes up and down for each rebound):$D = h + 2h_1 + 2h_2 + 2h_3 + \dots$$D = h + 2(e^2h + e^4h + e^6h + \dots)$$D = h + 2he^2(1 + e^2 + e^4 + \dots)$Using the sum of an infinite geometric series $S = \frac{a}{1-r}$ where $a = 1$ and $r = e^2$:$D = h + 2he^2 \left( \frac{1}{1 - e^2} \right)$$D = h \left( 1 + \frac{2e^2}{1 - e^2} \right) = h \left( \frac{1 - e^2 + 2e^2}{1 - e^2} \right) = h \left( \frac{1 + e^2}{1 - e^2} \right)$.

$A$ particle falls from a height $h$ upon a fixed horizontal plane and rebounds. If $e$ is the coefficient of restitution,the total distance travelled before rebounding has stopped is

Similar Questions

If $\overline{a}, \overline{b}, \overline{c}$ are non-coplanar unit vectors such that $\overline{a} \times (\overline{b} \times \overline{c}) = \frac{\overline{b} + \overline{c}}{\sqrt{2}}$,then the angle between $\overline{a}$ and $\overline{b}$ is

Ionisation enthalpies of $Ce$,$Pr$,and $Nd$ are higher than $Th$,$Pa$,and $U$. Why?

$A$ particle of mass $m$ is moving in a circular path of constant radius $r$ such that its centripetal acceleration $a_c$ is varying with time $t$ as $a_c = k^2rt^2$. The power delivered to the particle by the forces acting on it is

At a certain instant, a stationary transverse wave is found to have maximum kinetic energy. The appearance of the string at that instant is:

For a reaction $\frac{1}{2} A \rightarrow 2B,$ the rate of disappearance of $A$ is related to the rate of appearance of $B$ by the expression:

Vedclass Test Series

Exam Paper Generator

Online Exam Module