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 rolls the ball for the first time, (b) Find the speed of A after he catches the ball for the first time. (c) Find the speeds of A and B after the ball has made 5 round trips and is held by A. (d) how many times can A roll the ball? (e) Where is the center of mass of the system “A + B + ball” at the end of the nth trip?
Given: Weight of two friends A and B (each weighing 40 kg) is M are sitting on a frictionless platform some distance d apart.
A rolls a ball of mass (m) 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 (u) of 5 m/s on the platform.
{a} the speed (v) of A after he rolls the ball for the first time will be: Case1: Total momentum of the man A and the ball will remain constant
Therefore, mu - Mv = 0
0 = 4 x 5 - 40 x v
v = 0.5 m/s towards left
{b} the speed (v’) of A after he catches the ball for the first time will be:
Case2:
When B catches the ball, the momentum between the B and the ball will remain constant.
4 x 5 = 44
V1= 20/44 m/s
Case3: When B throws the ball, then applying Law of Conservation of Linear Momentum
44 x (20/44) = -4 x 5 + 40 x v2
20 = -20 + 40 v2
V2 = 40/40 m/s
V2 = 1 m/s towards right
Case 4: When A catches the ball, then applying Law of Conservation of Linear Momentum
Here (-v) = (-0.5) from case 1
-4 x 5 + (-0.5) x 40 = -44 v'
-20 - 20 = -44v'
v'= 10/11 m/s towards left
{c} the speeds of A and B after the ball has made 5 round trips and is held by A will be:
Case5: When A throws the ball, then applying Law of Conservation of Linear Momentum
44 x (10/11) = 4 x 5 - 40 x v4
v4 = 60/40
v4 = 3/2 m/s towards left
Case 6: When B receives the ball, then applying Law of Conservation of Linear Momentum
40 x 1 + 4 x 5 = 44 x v5
v5 = 60/44
v5 = 15/11 m/s towards right
Case7: When B throws the ball, then applying Law of Conservation of Linear Momentum
44 x (66/44) = -4 x 5 + 40 x v6
V6 = 80/40
V6 = 2 m/s towards right
Case 8: When A catches the ball, then applying Law of Conservation of Linear Momentum
-4 x 5 -40 x (3/2) = -44v'
V'= 80/44
v'= 20/11 m/s towards left
Similarly, after 5 round trips
Velocity of A will be 50/11 and velocity of B will be 5 m/s
{d} Since after 6th round trip, the velocity of A is
So, it can't catch the ball.
So, it can only roll the ball six.
{e} the center of mass of the system “A + B + ball” at the end of the nth trip will be:
Let the ball and the body A at the initial position is at origin.
Centre of mass of the system,
(At initial position v and u are both 0)
= 40d / 84
= 10d/11