Question

A ball is suspended by a thread from the ceiling of a car. The brakes are applied and the speed of the car changes uniformly from ${36kmh}^{-1}$ to zero in $5s$ . The angle by which the ball deviates from the vertical ($$g=10m{s}^{-2}$$) is:
(A) $$ta{n}^{-1}(\frac{1}{3})$$
(B) $$si{n}^{-1}(\frac{1}{5})$$
(C) $$ta{n}^{-1}(\frac{1}{5})$$
(D) $$co{t}^{-1}(\frac{1}{3})$$

Answer 1

Hint We will first find out the acceleration of the car. Finally we will find out the angle it makes with the vertical.
Formulae Used: $a=\frac{(v-u)}{t}$
Where, $a$ is the acceleration of the body, $v$ is the final velocity, $u$ is the initial velocity and $t$ is the time taken by the body.

Complete Step By Step Solution
Here,
$$u=36km{h}^{-1}=10m{s}^{-1}$$
$$v=0km{h}^{-1}=0m{s}^{-1}$$
$$t=5s$$
Now,
$$a_{car}=\frac{(0-10)}{5}$$
Now,
$$a_{}=-2m{s}^{-2}$$
$$\begin{array}{c}tan\theta =\frac{a}{g}=\frac{2}{10}=\frac{1}{5}\\ \end{array}$$
Thus, the answer turns out to be $$\theta =ta{n}^{-1}(\frac{1}{5})$$ which is (C).
But,
Also,
If we take
$$\begin{array}{c}sin\theta =\frac{2}{10}=\frac{1}{5}\\ \end{array}$$
$$\Rightarrow \theta =si{n}^{-1}(\frac{1}{5})$$

Hence, (B) is also correct.

Note We got the value of $$sin\theta$$ , we took $$a_{net}=\sqrt{a^2+g^2}$$ an d value turns out to be $$10m{s}^{-2}$$. Also, we took only the magnitude as it only plays the role to form the angle.