$$\text { Figure shows }(x, t),(y, t) \text { diagram of a particle moving in 2-dimensions. }$$
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 $$=\left(\frac{5}{3}+\frac{5}{3}\right) \mathrm{s}=\frac{10}{3} \mathrm{~s}=3.33 \mathrm{~s}$$
There are three forces $$\mathbf{F}_1, \mathbf{F}_2$$ and $$\mathbf{F}_3$$ acting on a body, all acting on a point $$P$$ on the body. The body is found to move with uniform speed.
(a) Show that the forces are coplanar.
(b) Show that the torque acting on the body about any point due to these three forces is zero.
As the body is moving with uniform speed (velocity) its acceleration $$a=0$$.
$$\therefore$$ The sum of the forces is zero, $$\mathrm{F}_1+\mathrm{F}_2+\mathrm{F}_3=0$$
(a) Let $$F_1, F_2, F_3$$ be the three forces passing through a point. Let $$F_1$$ and $$F_2$$ be in the plane $$A$$ (one can always draw a plane having two intersecting lines such that the two lines lie on the plane). Then $$F_1+F_2$$ must be in the plane $$A$$.
Since, $$F_3=-\left(F_1+F_2\right), F_3$$ is also in the plane $$A$$.
(b) Consider the torque of the forces about $$P$$. Since, all the forces pass through $$P$$, the torque is zero. Now, consider torque about another point $$O$$. Then torque about $$O$$ is
$$\begin{aligned} & \text { Torque }=\mathbf{O P} \times\left(F_1+F_2+F_3\right) \\ & \text { Since, } F_1+F_2+F_3=0 \text {, torque }=0 \end{aligned}$$
When a body slides down from rest along a smooth inclined plane making an angle of $$45^{\circ}$$ with the horizontal, it takes time $$T$$. When the same body slides down from rest along a rough inclined plane making the same angle and through the same distance, it is seen to take time $$p T$$, where $$p$$ is some number greater than 1. Calculate the coefficient of friction between the body and the rough plane.
Consider the diagram where a body slides down from along an inclined plane of inclination $$\theta\left(=45^{\circ}\right)$$.
On smooth inclined plane Acceleration of a body sliding down a smooth inclined plane
$$\begin{aligned} a & =g \sin \theta \\ \text{Here,}\quad\theta & =45^{\circ} \\ \therefore \quad a & =g \sin 45^{\circ}=\frac{g}{\sqrt{2}} \end{aligned}$$
Let the travelled distance be s.
Using equation of motion, $$s=u t+\frac{1}{2} a t^2$$, we get
$$\begin{aligned} & s=0 \cdot t+\frac{1}{2} \frac{g}{\sqrt{2}} T^2 \\ \text{or}\quad & s=\frac{g T^2}{2 \sqrt{2}}\quad \text{... (i)} \end{aligned}$$
On rough inclined plane Acceleration of the body $$a=g(\sin \theta-\mu \cos \theta)$$
$$\begin{aligned} & =g\left(\sin 45^{\circ}-\mu \cos 45^{\circ}\right) \\ & =\frac{g(1-\mu)}{\sqrt{2}} \quad\left(\text { As, } \sin 45^{\circ}=\cos 45^{\circ}=\frac{1}{\sqrt{2}}\right) \end{aligned}$$
Again using equation of motion, $$s=u t+\frac{1}{2} a t^2$$, we get
$$\begin{aligned} & s=0(p T)+\frac{1}{2} \frac{g(1-\mu)}{\sqrt{2}}(p T)^2 \\ \text{or}\quad & s=\frac{g(1-\mu) p^2 T^2}{2 \sqrt{2}} \quad \text{.... (ii)} \end{aligned}$$
From Eqs. (i) and (ii), we get
$$\frac{g T^2}{2 \sqrt{2}}=\frac{g(1-\mu) p^2 T^2}{2 \sqrt{2}}$$
or $$\quad (1-\mu) p^2=1$$
or $$\quad 1-\mu=\frac{1}{p^2}$$
or $$\quad \mu=\left(1-\frac{1}{p^2}\right)$$
Figure shows $$\left(v_x, t\right)$$, and $$\left(v_y, t\right)$$ diagrams for a body of unit mass. Find the force as a function of time.
Consider figure (a)
$$\begin{aligned} v_x & =2 t \text { for } 0 < t <1 \mathrm{~s} \\ & =2(2-t) \text { for } 1 < t < 2 \mathrm{~s} \\ & =0 \text { for } t>2 \mathrm{~s} \end{aligned}$$
From figure (b)
$$\begin{aligned} v_y & =t \text { for } 0 < t<1 \mathrm{~s} \\ & =1 \text { for } t>1 \mathrm{~s} \end{aligned}$$
$$\begin{aligned} \therefore \quad F_x & =m a_x=m \frac{d v_x}{d t} \quad \text{for } 0 < t < 1s\\ & =1 \times 2 \quad \text{for } 0 < t < 2s\\ & =1(-2) \quad \text{for } 2 < t \end{aligned}$$
$$\begin{aligned} F_y & =m a_y=m \frac{d v_y}{d t} \\ & =1 \times 1 \text { for } 0 < t <1 \mathrm{~s}=0 \text { for } 1 < t \\ \text{Hence,}\quad F & =F_x \hat{\mathbf{i}}+F_y \hat{\mathbf{j}} \quad \text{for } 0 < t < 1s\\ & =2 \hat{\mathbf{i}}+\hat{\mathbf{j}} \quad \text{for } 1 < t < 2s\\ & =-2 \hat{\mathbf{i}} \quad \text{for } t < 2s\\ & =0 \end{aligned}$$