Question

A small metal sphere falls from a height $h$ onto a plane surface strikes it with velocity $v$ and rebounds. If $e$ is the coefficient of restitution :
a) the height to which it rebounds after $n$ collisions is $e^n$ $h$
b) the height to which it rebound after 3 collisions is $e^6$ $h$
c) the velocity with which it rebounds after 4 collisions is $e^4$ $v$

d) the velocity with which it strikes the surface for the second collision is $ev$

• A. Only a and b are correct
• B. Only b and c are correct
• C. b , c and d are correct
• D. a, b and c are correct

Answer 1

The correct option is C b , c and d are correct

Let speed after $n^{th}$ collision be $v_n$ and height after $n^{th}$ collision be $h_n$.

Using conservation of mechanical energy,

$mg{h}_n=\frac{1}{2}m{v}_n^2$

$v_n=\sqrt{2g{h}_n}$

Using definition of coefficient of restitution,

$e=\frac{v_{n+1}}{v_n}$

After first collision, $v_1=ev$

Using above two, it can be concluded that, $v_n=e^{n}v$

Hence, (c) is correct.

$⟹\sqrt{2g{h}_n}=e^n\sqrt{2gh}$

$h_n=e^{2n}h$

Hence, (b) is correct.

From conservation of energy, velocity of sphere just before $(n+1{)}^{th}$ collision is same as the velocity of sphere just after $n^{th}$ collision.

Hence, (d) is correct.