Question

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. 0.16
• B. 0.32
• C. 0.4
• D. 0.56

Answer 1

The correct option is C 0.4

Ball is dropped form height h = 25 cm = $\frac{1}{4}m$
$\Rightarrow$ Velocity when it strikes ground is,
$u=\sqrt{2gh}=\sqrt{2\times 10\times \frac{1}{4}}=\sqrt{5}m/s$



Let the velocity acquired immediately after collision is V in upward direction.

$\Rightarrow {V_f}^2=V^2+2as$

$or,O=V^2-2g\left(\frac{4}{100}\right)$

$or,V=\sqrt{0.8}m/s$

$e=\frac{speedofseparation}{speedofapproach}$

$\Rightarrow e=\frac{V}{u}$

$\Rightarrow e=\sqrt{\frac{0.8}{5}}=\sqrt{\frac{4}{25}}=\frac{2}{5}$

$\therefore e=0.4$

Why this question? Tip: We can either use mechanical energy conservation or 3rd equation of motion to find u.