Question 341: Work, Energy and Power & Circular Motion

Question

There is a rectangular plate of mass M kg of dimensions ( $a 	imes b$ ). The plate is held in horizontal position by striking $n$ small balls each of mass m per unit area per unit time. These are striking in the shaded half region of the plate. The balls are colliding elastically with velocity $v$. What is $v$ ?
It is given $n=100, \mathrm{M}=3 \mathrm{~kg}, m=0.01 \mathrm{~kg}$; $b=2 m ; a=1 \mathrm{~m} ; g=10 \mathrm{~m} / \mathrm{s}^2$

Accepted Answer

The ball collides elastically with the stationary plate. In this collision, velocity of the ball gets reversed.
The conversation of linear momentum,
$$ \begin{aligned} \mathrm{P}_{i(	ext { ball })}+\mathrm{P}_{i(	ext { plate })} & =\mathrm{P}_{f(	ext { ball })}+\mathrm{P}_{f(	ext { plate })} \ \Delta \mathrm{P}_{	ext {plate }} & =-\Delta \mathrm{P}_{	ext {ball }} \ & =-m v-(-m v) \ & =2 m v 	ext { (upwards). } \end{aligned} $$
number of balls striking the plate in times $\Delta t$ is $\mathrm{N}=n(a b / 2) \Delta \mathrm{t}=n a b \Delta t / 2$.
Thus, the total change in the plate's momentum in time $\Delta t$ is
$$ \Delta \mathrm{P}=\mathrm{N} \Delta \mathrm{P}_{(	ext {plate })}=	ext { mvnab } \Delta t . $$
By Newton's second law, upward force on the
$$ 	ext { plate is } \mathrm{F}=\frac{\Delta \mathrm{P}}{\Delta t}=m v n a b .\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i) $$
This force effectively acts at the centre of the shaded region.
i.e., at a distance $\frac{3 b}{4}$ from the hinge.
In equillibrium $\mathrm{F}+\mathrm{R}=m g$, and torque due to all the forces about any point is zero.
The torque about a point on the hinge is
$$ \mathrm{F}\left(\frac{3 b}{4}ight)-\operatorname{Mg}\left(\frac{b}{2}ight)=0 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)$$
Substitute value of F from equation (i) into equation (ii) to get
$$ \begin{aligned} \mathrm{F} & =\frac{\mathrm{Mg} b}{2} 	imes \frac{4}{3 b}=\frac{\mathrm{Mg} 2}{3}=	ext { mvnab } \ v & =\frac{q 2 \mathrm{Mg}}{3 m n a b}=\frac{2 	imes 3 	imes 10}{3 	imes 0.01 	imes 100 	imes 1 	imes 2} \ & =\frac{60}{6}=10 \mathrm{~m} / \mathrm{s} \end{aligned} $$

Work, Energy and Power & Circular Motion

Integer / Numeric Answer Type