Question

Two friends A and B (each weighing 40 kg) are sitting on a frictionless platform some distance d apart. A rolls a ball of mass 4 kg on the platform towards B which B catches. Then B rolls the ball towards A and A catches it. The ball keeps on moving back and forth between A and B. The ball has a fixed speed of 5 m/s on the platform. (a) Find the speed of A after he catches the ball for the first time. (c) Find the speeds of A and B after the all has made 5 round trips and is held by A. (d) How many times can A roll the ball? (e) Where is the centre of mass of the system "A + B + ball" at the end of the nth trip?

Answer 1

It is given that:
Weight of A = Weight of B = 40 kg
Velocity of ball = 5 m/s

(a) Case-1: Total momentum of the man A and ball remains constant.

∴ 0 = 4 × 5 − 40 × v
⇒ v = 0.5 m/s, towards left

(b) Case-2: When B catches the ball, the momentum between B and the ball remains constant.

⇒ 4 × 5 = 44 v
⇒ $v=\left(\frac{20}{44}\right)\mathrm{m}/\mathrm{s}$

Case-3: When B throws the ball,

On applying the law of conservation of linear momentum, we get:

$$\begin{gathered} \Rightarrow 44\times \left(\frac{20}{44}\right)=-4\times 5+40\times v \\ \Rightarrow v=1\mathrm{m}/\mathrm{s},(\mathrm{towards}\mathrm{right}) \end{gathered}$$

Case-4: When A catches the ball,

Applying the law of conservation of liner momentum, we get:

$$\begin{gathered} -4\times 5+(-0.5)\times 40=44v \\ \Rightarrow v=\frac{40}{44}=\frac{10}{11}\mathrm{m}/\mathrm{s},\mathrm{towards}\mathrm{left} \end{gathered}$$


(c) Case-5: When A throws the ball,

Applying the law of conservation of linear momentum, we get:

$$\begin{gathered} 44\times \left(\frac{10}{11}\right)=4\times 5+40\times v \\ \Rightarrow v=\frac{60}{40}=\frac{3}{2}\mathrm{m}/\mathrm{s}(\mathrm{towards}\mathrm{left}) \end{gathered}$$

Case-6: When B receives the ball,

Applying the law of conservation of linear momentum, we get:

$$\begin{gathered} 40\times 1+4\times 5=44\times v \\ \Rightarrow v=\frac{60}{44}\mathrm{m}/\mathrm{s},\mathrm{towards}\mathrm{right} \end{gathered}$$

Case-7: When B throws the ball,

On applying the law of conservation of linear momentum, we get:

$\Rightarrow v=44\times \left(\frac{60}{44}\right)\mathrm{m}/\mathrm{s},\mathrm{towards}\mathrm{right}$

Case-8: When A catches the ball,

On applying the law of conservation of linear momentum, we get:
$$\begin{gathered} -4\times 5+40\left(\frac{3}{2}\right)=-44v \\ \Rightarrow \mathrm{V}=-\frac{40}{44}=\left({\displaystyle \frac{10}{11}}\right)\mathrm{m}/\mathrm{s},\mathrm{towards}\mathrm{left} \end{gathered}$$

Similarly, after 5 round trips,
The velocity of A will be $\left(\frac{50}{11}\right)$ m/s and the velocity of B will be 5 m/s.

(d) As after 6 round trips, the velocity of A becomes $\frac{60}{11}$ i.e. > 5 m/s, it cannot catch the ball. Thus, A can only roll the ball six times.

(e) Let the ball and the body A be at origin, in the initial position.

$$\begin{gathered} \therefore {\mathrm{X}}_{\mathrm{c}}=\frac{40\times 0+4\times 0+40\times d}{40+40+4} \\ =\frac{10}{21}d \end{gathered}$$