A ball is projected from a point in one of the two smooth parallel vertical walls against the other in a plane perpendicular to both. After being reflected at each wall impinge again on the second at a point in the same plane as it started. The distance between two walls is $a . b$ is the free range and $e$ be the coefficient of restitution:

The total time taken in moving from

O

C

\frac{a}{e^{2} u} (e^{2} + e + 1)

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

The free range on the horizontal plane

b = \frac{2 u v}{g}

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

b e^{2} = a (e^{2} + e + 1)

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

All of these

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

Open in App

Solution

The correct options are
A The total time taken in moving from $O$ to $C$ is $\frac{a}{e^{2} u} (e^{2} + e + 1)$
B The free range on the horizontal plane $b = \frac{2 u v}{g}$
C $b e^{2} = a (e^{2} + e + 1)$
D All of these
$R = \frac{2 u_{x} u_{y}}{g} = \frac{2 u v}{g} = b$
$t = \frac{a}{u} + \frac{a}{e u} + \frac{a}{e^{2} u} = \frac{a}{u} (1 + \frac{1}{e} + \frac{1}{e^{2}})$
$= \frac{a}{u} (\frac{e^{2} + e + 1}{e^{2}}) = \frac{a}{e^{2} u} (e^{2} + e + 1)$
$∴ \frac{2 v}{g} = \frac{a}{e^{2} u} (e^{2} + e + 1)$
$∴ b e^{2} = a (e^{2} + e + 1)$

Suggest Corrections

Similar questions

Q. A ball is projected from a point in one of the two smooth parallel vertical walls against the other in a plane perpendicular to both. After being reflected at each wall impinge again on the second at a point in the same plane as it started. The distance between two walls is

a . b

is the free range and

e

be the coefficient of restitution:

Q. A ball is projected from a point in a horizontal plane so as to strike a vertical wall at right angle and after rebounding from the wall and once from the horizontal plane it returns to the point of projection. Find coefficient of restitution (e) for the two collisions, assuming it to be same for both collision. Find the value of 1/e.

A ball is projected with a velocity $u$ at an elevation from a point distance d from a smooth vertical wall in a plane perpendicular to it. After rebounding from the wall, it returns to the point of projection, find the maximum distance $d$ for which the ball can return to the point of projection.

Q. A ball is projected with a velocity

u

at an elevation from a point distance

d

from a smooth vertical wall in a plane perpendicular to it. After rebounding from the wall, it returns to the point projection, proves that

u^{2} s i n 2 α = g f (1 + 1 / e)

. Hence find the maximum distance

d

for which the ball can return to the point of projection.

Join BYJU'S Learning Program

The correct options are A The total time taken in moving from O to C is ae2u(e2+e+1) B The free range on the horizontal plane b=2uvg C be2=a(e2+e+1) D All of theseR=2uxuyg=2uvg=b t=au+aeu+ae2u=au(1+1e+1e2) =au(e2+e+1e2)=ae2u(e2+e+1) ∴2vg=ae2u(e2+e+1) ∴be2=a(e2+e+1)