A small ball moving with a velocity 10 m/s, horizontally (as shown in figure) strikes a rough horizontal surface having $μ = 0 .5$ . If the coefficient of restitution is e = 0.4. Horizontal component of velocity of ball after 1^st impact will be $(g = 10 m / s^{2})$

see full answer

21% of IItians & 23% of AIIMS delhi doctors are from Sri Chaitanya institute !!

An Intiative by Sri Chaitanya

answer is 3.

(Unlock A.I Detailed Solution for FREE)

Ready to Test Your Skills?

Check your Performance Today with our Free Mock Test used by Toppers!

Take Free Test

Detailed Solution

Vertical component of velocity after impact, $e = \frac{v_{y}}{u_{y}} \Rightarrow v_{y} = 0 .4 \times 10 = 4 m / s$

In vertical direction, impulse-momentum relation is

$\int Ndt = {mv}_{y} - {mu}_{y}$

$\Rightarrow \int Ndt = m \times 4 - (- 10 m) = 14 m$

In horizontal direction, impulse-momentum relation is

$- \int μNdt = {mv}_{x} - {mu}_{x}$

$\Rightarrow - 0 .5 \times 14 m = {mv}_{x} - m \times 10$

$\Rightarrow v_{x} = 3 m / s$

Watch 3-min video & get full concept clarity