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 Bafter 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
⇒ \[v = \left( \frac{20}{44} \right)\text{ m/s}\]
Case-3: When B throws the ball,
On applying the law of conservation of linear momentum, we get:

\[\Rightarrow 44 \times \left( \frac{20}{44} \right) = - 4 \times 5 + 40 \times v\]

\[ \Rightarrow v = 1 \text{ m/s , towards right)}\]

Case-4: When A catches the ball,

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

\[- 4 \times 5 + ( - 0 . 5) \times 40 = 44 v\]

\[ \Rightarrow v = \frac{40}{44} = \frac{10}{11}\text{ m/s, towards left}\]
(c) Case-5: When A throws the ball,

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

\[44 \times \left( \frac{10}{11} \right) = 4 \times 5 + 40 \times v\]

\[ \Rightarrow v = \frac{60}{40} = \frac{3}{2} \text{ m/s towards left)}\]

Case-6: When B receives the ball,

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

\[40 \times 1 + 4 \times 5 = 44 \times v\]

\[ \Rightarrow v = \frac{60}{44} \text{ m/s, towards right }\]

Case-7: When B throws the ball,

On applying the law of conservation of linear momentum, we get:
\[\Rightarrow v = 44 \times \left( \frac{60}{44} \right) \text{ m/s, towards right }\]
Case-8: When A catches the ball,

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

\[- 4 \times 5 + 40\left( \frac{3}{2} \right) = - 44v\]

\[ \Rightarrow V = - \frac{40}{44} = $\left( \frac{10}{11} \right) \text{ m/s,towards left } \]

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 \[\frac{60}{11}\] > 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.

\[\therefore X_c = \frac{40 \times 0 + 4 \times 0 + 40 \times d}{40 + 40 + 4}\]

\[ = \frac{10}{21}d\]