Block $$A$$ of weight 100 N rests on a frictionless inclined plane of slope angle $$30^{\circ}$$. A flexible cord attached to $$A$$ passes over a frictionless pulley and is connected to block $$B$$ of weight $$w$$. Find the weight $$w$$ for which the system is in equilibrium.
In equilibrium, the force $$m g \sin \theta$$ acting on block $$A$$ parallel to the plane should be balanced by the tension in the string, i.e.,
$$\begin{aligned} m g \sin \theta & =T=F \quad [\because \text{T=F given}]\quad \text{.... (i)}\\ w & =T=F \quad \text{... (ii)} \end{aligned}$$
and for block B, where, $$w$$ is the weight of block $$B$$.
From Eqs. (i) and (ii), we get,
$$\begin{aligned} \therefore \quad W & =m g \sin \theta \\ & =100 \times \sin 30^{\circ} \quad (\because m g=100 \mathrm{~N})\\ & =100 \times \frac{1}{2} \mathrm{~N}=50 \mathrm{~N} \end{aligned}$$
A block of mass $$M$$ is held against a rough vertical wall by pressing it with a finger. If the coefficient of friction between the block and the wall is $$\mu$$ and the acceleration due to gravity is $$g$$, calculate the minimum force required to be applied by the finger to hold the block against the wall.
Given, mass of the block $$=M$$
Coefficient of friction between the block and the wall $$=\mu$$
Let a force $$F$$ be applied on the block to hold the block against the wall. The normal reaction of mass be $$N$$ and force of friction acting upward be $$f$$. In equilibrium, vertical and horizontal forces should be balanced separately.
$$\begin{array}{ll} \therefore & f=M g \quad \text{... (i)}\\ \text { and } & F=N \quad \text{... (ii)} \end{array}$$
$$\text { But force of friction }(f)=\mu \mathrm{N}$$
$$=\mu F\quad \text{[using Eq. (ii)] .... (iii)}$$
$$\begin{aligned} &\text { From Eqs. (i) and (iii), we get } \quad \mu F =M g \\ &\text { or } \quad F =\frac{M g}{\mu} \\ &\begin{aligned} \\ \end{aligned} \end{aligned}$$
A 100 kg gun fires a ball of 1 kg horizontally from a cliff of height 500 m. It falls on the ground at a distance of 400 m from the bottom of the cliff. Find the recoil velocity of the gun. (acceleration due to gravity $$=10 \mathrm{~ms}^{-2}$$)
$$\text { Given, mass of the gun }\left(m_1\right)=100 \mathrm{~kg}$$
Mass of the ball $$\left(m_2\right)=1 \mathrm{~kg}$$
Height of the cliff $$(h)=500 \mathrm{~m}$$
Horizontal distance travelled by the ball $$(x)=400 \mathrm{~m}$$
From $$h=\frac{1}{2} g t^2 \quad(\because \text { Initial velocity in downward direction is zero)}$$
$$\begin{aligned} 500 & =\frac{1}{2} \times 10 t^2 \\ t & =\sqrt{100}=10 \mathrm{~s} \end{aligned}$$
$$\text { From } \quad x=u t, u=\frac{x}{t}=\frac{400}{10}=40 \mathrm{~m} / \mathrm{s}$$
$$\begin{aligned} &\text { If } v \text { is recoil velocity of gun, then according to principle of conservation of linear momentum, }\\ &\begin{aligned} m_1 v & =m_2 u \\ v & =\frac{m_2 u}{m_1}=\frac{1}{100} \times 40=0.4 \mathrm{~m} / \mathrm{s} \end{aligned} \end{aligned}$$
$$\text { Figure shows }(x, t),(y, t) \text { diagram of a particle moving in 2-dimensions. }$$
If the particle has a mass of 500 g, find the force (direction and magnitude) acting on the particle.
Clearly from diagram (a), the variation can be related as
$$\begin{aligned} x & =t \Rightarrow \frac{d x}{d t}=1 \mathrm{~m} / \mathrm{s} \\ a_x & =0 \end{aligned}$$
$$\begin{aligned} &\text { From diagram (b) }\\ &y=t^2 \end{aligned}$$
$$\Rightarrow \quad \frac{d y}{d t}=2 t \text { or } a_y=\frac{d^2 y}{d t^2}=2 \mathrm{~m} / \mathrm{s}^2$$
$$\text { Hence, } \quad I_y=m a_y=500 \times 10^{-3} \times 2=1 \mathrm{~N} \quad(\because m=500 \mathrm{~g})$$
$$F_x=m a_x=0$$
Hence, net force, $$F=\sqrt{F_x^2+F_y^2}=F_y=1 \mathrm{~N}\quad$$ (along $$y$$-axis)
A person in an elevator accelerating upwards with an acceleration of $$2 \mathrm{~ms}^{-2}$$, tosses a coin vertically upwards with a speed of $$20 \mathrm{~ms}^{-1}$$. After how much time will the coin fall back into his hand? $$\left(g=10 \mathrm{~ms}^{-2}\right)$$
Here, initial speed of the $$\operatorname{coin}(u)=20 \mathrm{~m} / \mathrm{s}$$
Acceleration of the elevator (a) $$=2 \mathrm{~m} / \mathrm{s}^2\quad$$ (upwards)
Acceleration due to gravity $$(g)=10 \mathrm{~m} / \mathrm{s}^2$$
$$\therefore \text { Effective acceleration } a^{\prime}=g+a=10+2=12 \mathrm{~m} / \mathrm{s}^2 \quad \text { (here, acceleration is w.r.t. the lift) }$$
$$\text { If the time of ascent of the coin is } t \text {, then }$$
$$\begin{aligned} & v=u+a t \\ & 0=20+(-12) \times t \\ \text{or}\quad & t=\frac{20}{12}=\frac{5}{3} \mathrm{~s} \end{aligned}$$
Time of ascent $$=$$ Time of desent
$$\therefore \quad$$ Total time after which the coin fall back into hand $$=\left(\frac{5}{3}+\frac{5}{3}\right) \mathrm{s}=\frac{10}{3} \mathrm{~s}=3.33 \mathrm{~s}$$