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

Explanation:

a) CASE-1:- Man A's overall momentum and the ball will stay constant

Consequently,

0 = 4 x 5 - 40 x v

v = 0.5 m/s to the left  

(b) CASE - 2  

Whether B catches the ball, the momentum remains constant between the B and the ball.

4 x 5 = 44v

$\mathrm{v}=\frac{20}{44} \mathrm{m} / \mathrm{s}$

CASE - 3 => If B throws the ball, the Linear Momentum Conservation Law (L.C.L.M.) applies.

44 x$\left(\frac{20}{44}\right)$ = -4 x 5 + 40 x v

20 = -20 + 40v

v = $\frac{40}{40}$ = 1 m/s to the right

CASE - 4 => When A picks up the ball, apply L.C.L.M.

-4 x 5 + (-0.5) x 40 = -44 v

-20 - 20 = -44 v

v =$\frac{40}{44}$$m/s$=$\frac{10}{11}$$m / s$ to the  left

(c) CASE - 5  

If A throws the ball, use L.C.L.M.

44 x $\left(\frac{10}{11}\right)$ = 4 x 5 - 40 x v

$v=\frac{60}{40}$

$\mathrm{v}=\frac{3}{2} \mathrm{m} / \mathrm{s}$ towards left

CASE - 6  

When the ball is received by B, apply L.C.L.M.

40 x 1 + 4 x 5 = 44 x v

$v=\frac{60}{44}$

$\mathrm{v}=\frac{15}{11} \mathrm{m} / \mathrm{s}$

CASE - 7  

If B throws the ball, use L.C.L.M..

$44 \times\left(\frac{66}{44}\right)=-4 \times 5+40 x$

$v=\frac{\beta 0}{40}$

v = 2 m/s to the right

CASE - 8  

When A picks up the ball, apply L.C.L.M.

$\begin{aligned}&-4 \times 5-40 \times\left(\frac{3}{2}\right)=-44 v\\&v=\frac{80}{44}\end{aligned}$

$\mathrm{N}=\frac{20}{11} \mathrm{m} / \mathrm{s}$ towards left

Likewise after 5 round trips A speed will be 50/11 and B speed 5 m / s

d) Since the speed of A is 60/11 m / s > 5 m / s after the 6th round trip

So, the ball can't catch it.

So, it can just roll the sixth ball.

(e) Let the ball and body A originate in the initial position.

The system's mass center $=\frac{(40 \times 0+4 \times 0+40 \times d)}{(40+40+4)}$

$\begin{aligned}&=\frac{40 \mathrm{d}}{(84)}\\&=\frac{10 d}{(21)}\end{aligned}$