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 questions:

What is the speed of $A$ when it hits $B$?

• A. $\frac{28}{5}m{s}^{-1}$
• B. $\frac{56}{5\sqrt{3}}m{s}^{-1}$
• C. $\frac{56}{5}m{s}^{-1}$
• D. None of these

Answer 1

The correct option is A

$\frac{28}{5}m{s}^{-1}$

Step 1: Given data

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

The angle of projection, $\theta =37°$

The initial height of ball, $B$$=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, ${\mathit{S}}_{\mathbf{y}}\mathbf{}\mathit{o}\mathit{f}\mathbf{}\mathit{A}\mathbf{=}{\mathit{S}}_{\mathbf{y}}\mathbf{}\mathit{o}\mathit{f}\mathbf{}\mathit{B}$ at $\mathit{t}\mathbf{=}\mathbf{}\frac{\mathbf{3}}{\mathbf{7}}\mathbf{}\mathit{s}$

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,

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

The $y$ component of the speed of the ball at $t=\frac{3}{7}s$ is $\mathbf{0}\mathbf{}\mathit{m}{\mathit{s}}^{\mathbf{-}\mathbf{1}}$.

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 at $t=\frac{3}{7}s$ is $\frac{\mathbf{28}}{\mathbf{5}}\mathbf{}\mathit{m}$.

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

Speed of the ball $A$ at $t=\frac{3}{7}s$, $\left|\stackrel{\rightharpoonup }{u}\right|=\sqrt{{\left(u_x\right)}^2+{\left(u_y\right)}^2}$

$$\begin{gathered} =\sqrt{{\left(\frac{28}{5}\right)}^2+0} \\ =\frac{28}{5}m{s}^{-1} \end{gathered}$$

Thus, the speed of ball $\mathit{A}$ when it hits the ball $\mathit{B}$ is $\frac{\mathbf{28}}{\mathbf{5}}\mathbf{}\mathit{m}{\mathit{s}}^{\mathbf{-}\mathbf{1}}$. Hence. option A is correct.