Question

a 100g ball having speed 10m/s collides with a wall and rebounds with the same speed and the contact period between the ball and wall is 0.1 second, then the force experienced by the wall is

Answer 1

${\mathfrak{\red{\huge{\underline{Answer}}}}}$

$<b><I>As we know that$
F = mass × acceleration
Given
mass = 100g
u = 10m/s
t = 0.1
and acceleration = (v-u)/t
now
given that |u| = |v| = 10m/s
but direction is opposite as it rebounds
then
u =10 m/s
v =-10 m/s

then acceleration
$= \frac{ (- 10) - 10}{ \frac{1}{10} } \\ \\ = \frac{ - 20}{1} \times 10 \\ \\ = - 200 \frac{m}{ {s}^{2} }$
now as
a = -200m/s^2

then force
$= \frac{100}{1000} \times - 200 \\ \\ = - \frac{ - 20000}{1000} \\ \\ = - 20 \: Newton$

Or force experience by wall
= 20 Newton
In magnitude

${\red{\boxed{\mathcal{\green{Hope\: it \:helps}}}}}$

Answer 2

Answer:

Explanation

Here in

This question angle is also given i. e 30° .

Here vertical momentum doesn't change bt horizontal momentum changes by sign.

F= change in momentum/ t

F= 2 mvcos60°

F= 2×100×10×10^-3/2×0.1

F=10N