Question

A ball $A$ is projected from $O$ with an initial velocity $v_0=7m{s}^{-1}$ in a direction $37°$ above the horizontal. Another ball $B$, $3m$ from $O$ on a line $37°$ above the horizontal is released from rest at the instant $A$ starts, as shown in figure.

[Take $\sin 37°=\frac{3}{7};\cos 37°=\frac{4}{5};g=9.8m{s}^{-2}$]

Based on above information, answer the following question:

In which direction $A$ is moving when it hits $B$?

• A. $0°$
• B. $30°$ upward
• C. $30°$ downward
• D. Can't be predicted

Answer 1

The correct option is A

$0°$

Step 1: Given data

The initial velocity of ball $A$, $v_0=7m{s}^{-1}$

Angle of projection $\theta =37°$

The distance of the ball $B$ from the origin is$3m$.

Step 2: Calculate the time of the collision

From the above figure in the right-angled triangle$∆BCA$:

$$\begin{gathered} \cos 37°=\frac{OC}{AB} \\ \Rightarrow OC=AB\times \cos 37° \\ \Rightarrow OC=3\times \frac{4}{5} \\ \Rightarrow OC=\frac{12}{5}m \end{gathered}$$

As the ball $B$ is moving in a vertically downward motion, it means that in order to collide with $B$ the horizontal displacement of $A$ will be $OC$ at time of collision $t$.

Using the third equation of motion under the gravity in the horizontal direction ${\mathit{s}}_{\mathbf{x}}\mathbf{=}{\mathit{u}}_{\mathbf{x}}\mathit{t}$ we get,

$$\begin{gathered} \Rightarrow OC=v_0\cos 37°\times t \\ \Rightarrow \frac{12}{5}=7\times \frac{4}{5}\times t \\ \Rightarrow t=\frac{12\times 5}{5\times 7\times 4} \\ \Rightarrow t=\frac{3}{7}s \end{gathered}$$

Step 3: Calculate the y component of the ball when the ball $\mathit{A}$ hits $\mathit{B}$

The height of both the balls would be the same when both the balls collided. This means the $y$ component of the displacement of both the balls will be the same at the time of the collision.

Therefore, $S_{y}ofA=S_{y}ofB$at $t=\frac{3}{7}s$

Using the third equation of motion under the gravity ${\mathit{s}}_{\mathbf{y}}\mathbf{=}{\mathit{u}}_{\mathbf{y}}\mathit{t}\mathbf{-}\frac{\mathbf{1}}{\mathbf{2}}{\mathit{g}}_{\mathbf{y}}{\mathit{t}}^{\mathbf{2}}$ we get,

$$\begin{gathered} \Rightarrow u_{A}t-\frac{1}{2}g{t}^2=u_{B}t-\frac{1}{2}g{t}^2 \\ \Rightarrow u_A\times \frac{3}{7}-\frac{g}{2}{\left(\frac{3}{7}\right)}^2=-\frac{g}{2}{\left(\frac{3}{7}\right)}^2 \\ \Rightarrow u_A\times \frac{3}{7}-\frac{g}{2}{\left(\frac{3}{7}\right)}^2=-\frac{g}{2}{\left(\frac{3}{7}\right)}^2 \\ \Rightarrow u_A\times \frac{3}{7}=0 \\ \Rightarrow u_A=0 \end{gathered}$$

The $y$ component of the speed of the ball $A$ at $t=\frac{3}{7}s$ is zero. Therefore, ${\mathit{u}}_{\mathbf{y}}\mathbf{=}\mathbf{0}$

Step 4: Calculate the x component of the ball when the ball $\mathit{A}$ hits $\mathit{B}$

The horizontal component of speed always remains constant throughout the projectile motion.

$$\begin{gathered} \therefore u_x=u\cos \theta \\ =7\times \cos 37° \\ =7\times \frac{4}{5} \\ =\frac{28}{5}m \end{gathered}$$

The $x$ component of the speed of the ball $A$ at $t=\frac{3}{7}s$ is $\frac{28}{5}m$

Step 5: Calculate the direction of the ball $\mathit{A}$ at $\mathit{t}\mathbf{=}\mathbf{}\frac{\mathbf{3}}{\mathbf{7}}\mathbf{}\mathit{s}$

The $y$ component of the speed of the ball and the $x$ component of the speed of the ball $A$ are perpendicular to each other.

$\tan \theta =\frac{u_y}{u_x}=\frac{0}{{\displaystyle \frac{28}{5}}}=0$ at $t=\frac{3}{7}s$

$$\begin{gathered} \Rightarrow \tan \theta =0 \\ \Rightarrow \theta ={\tan }^{-1}\left(0\right) \\ \Rightarrow \theta =0° \end{gathered}$$

Thus, the ball $A$ was at $0°$ with the positive $x-axis$ when it hits the ball $B$.

Hence, option A is correct.