A cricket bowler releases the ball in two different ways
(a) giving it only horizontal velocity, and
(b) giving it horizontal velocity and a small downward velocity.
The speed $$v_s$$ at the time of release is the same. Both are released at a height $$H$$ from the ground. Which one will have greater speed when the ball hits the ground? Neglect air resistance.
(a) When ball is given only horizontal velocity Horizontal velocity at the time of release $$\left(u_x\right)=v_s$$
During projectile motion, horizontal velocity remains unchanged,
$$\begin{aligned} \text{Therefore, } \quad & v_x=u_x=v_s \\ \text{In vertical direction,} \quad & v_y^2=u_y^2+2 g H \\ v_y=\sqrt{2 g H} \quad \left(\because u_y=0\right) \end{aligned}$$
$$\begin{aligned} &\therefore \text { Resultant speed of the ball at bottom, }\\ &\begin{aligned} v & =\sqrt{v_x^2+v_y^2} \\ & =\sqrt{v_s^2+2 g H} \quad \text{... (i)} \end{aligned} \end{aligned}$$
$$\text { (b) When ball is given horizontal velocity and a small downward velocity }$$
Let the ball be given a small downward velocity $$u$$.
$$\begin{aligned} \text{In horizontal direction} \quad & v_x^{\prime}=u_x=v_s \\ \text{In vertical direction } \quad & v_y^{\prime 2}=u^2+2 g H \\ \text{or} \quad & v_y^{\prime}=\sqrt{u^2+2 g H} \end{aligned}$$
$$\therefore$$ Resultant speed of the ball at the bottom
$$v^{\prime}=\sqrt{v_x^{\prime 2}+v_y^{\prime 2}}=\sqrt{v_s^2+u^2+2 g H} \quad \text{... (ii)}$$
From Eqs. (i) and (ii), we get $$\quad v^{\prime}>v$$
There are four forces acting at a point $$P$$ produced by strings as shown in figure. Which is at rest? Find the forces $$\mathbf{F}_1$$ and $$\mathbf{F}_2$$.
Consider the adjacent diagram, in which forces are resolved.
On resolving forces into rectangular components, in equilibrium forces $$\left(F_1+\frac{1}{\sqrt{2}}\right) N$$ are equal to $$\sqrt{2} N$$ and $$\boldsymbol{F}_2$$ is equal to $$\left(\sqrt{2}+\frac{1}{\sqrt{2}}\right) \mathrm{N}$$.
$$\begin{array}{rlrl} \therefore & F_1+\frac{1}{\sqrt{2}} & =\sqrt{2} \\ & & F_1 & =\sqrt{2}-\frac{1}{\sqrt{2}}=\frac{2-1}{\sqrt{2}}=\frac{1}{\sqrt{2}}=0.707 \mathrm{~N} \\ & \text { and } & F_2 & =\sqrt{2}+\frac{1}{\sqrt{2}}=\frac{2+1}{\sqrt{2}} \mathrm{~N}=\frac{3}{\sqrt{2}} \mathrm{~N}=2.121 \mathrm{~N} \end{array}$$
A rectangular box lies on a rough inclined surface. The coefficient of friction between the surface and the box is $$\mu$$. Let the mass of the box be $$m$$.
(a) At what angle of inclination $$\theta$$ of the plane to the horizontal will the box just start to slide down the plane?
(b) What is the force acting on the box down the plane, if the angle of inclination of the plane is increased to $$\alpha>\theta$$ ?
(c) What is the force needed to be applied upwards along the plane to make the box either remain stationary or just move up with uniform speed?
(d) What is the force needed to be applied upwards along the plane to make the box move up the plane with acceleration $$a$$ ?
(a) Consider the adjacent diagram, force of friction on the box will act up the plane.
$$\begin{aligned} &\text { For the box to just starts sliding down } m g\\ &\begin{array}{ll} & \sin \theta=f=\mu N=\mu m g \cos \theta \\ \text { or } \quad & \tan \theta=\mu \Rightarrow \theta=\tan ^{-1}(\mu) \end{array} \end{aligned}$$
(b) When angle of inclination is increased to $$\alpha>\theta$$, then net force acting on the box, down the plane is
$$\begin{aligned} F_1 & =m g \sin \alpha-f=m g \sin \alpha-\mu N \\ & =m g(\sin \alpha-\mu \cos \alpha) . \end{aligned}$$
(c) To keep the box either stationary or just move it up with uniform speed, upward force needed, $$F_2=m g \sin \alpha+f=m g(\sin \alpha+\mu \cos \alpha)$$ (In this case, friction would act down the plane).
(d) If the box is to be moved with an upward acceleration $$a$$, then upward force needed, $$F_3=m g(\sin \alpha+\mu \cos \alpha)+m a$$.
A helicopter of mass 2000 kg rises with a vertical acceleration of $$15 \mathrm{~ms}^{-2}$$. The total mass of the crew and passengers is 500 kg . Give the magnitude and direction of the $$\left(g=10 \mathrm{~ms}^{-2}\right)$$
(a) force on the floor of the helicopter by the crew and passengers.
(b) action of the rotor of the helicopter on the surrounding air.
(c) force on the helicopter due to the surrounding air.
Given, mass of helicopter $$\left(m_1\right)=2000 \mathrm{~kg}$$
Mass of the crew and passengeres $$m_2=500 \mathrm{~kg}$$
Acceleration in vertical direction $$a=15 \mathrm{~m} / \mathrm{s}^2(\uparrow)$$ and $$g=10 \mathrm{~m} / \mathrm{s}^2(\downarrow)$$
(a) Force on the floor of the helicopter by the crew and passengers
$$\begin{aligned} & m_2(g+a)=500(10+15) \mathrm{N} \\ & 500 \times 25 \mathrm{~N}=12500 \mathrm{~N} \end{aligned}$$
(b) Action of the rotor of the helicopter on the surrounding air $$=\left(m_1+m_2\right)(g+a)$$
$$\begin{aligned} & =(2000+500) \times(10+15)=2500 \times 25 \\ & =62500 \mathrm{~N}(\text { downward }) \end{aligned}$$
(c) Force on the helicopter due to the surrounding air
$$\begin{aligned} & =\text { reaction of force applied by helicopter. } \\ & =62500 \mathrm{~N} \text { (upward) } \end{aligned}$$