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?
Answers
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ANSWER::
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) CASE - 1 => Total momentum of the man A and the ball will remain constant
Therefore ,
0 = 4 x 5 - 40 x v
v = 0.5 m/s towards left
(b) CASE - 2 => When B catches the ball , the momentum between the B and the ball will remain constant.
4 x 5 = 44v
v = 20/44 m/s
CASE - 3 => When B throws the ball , then applying Law of Conservation of Linear Momentum (L.C.L.M.)
44 x (20/44) = -4 x 5 + 40 x v
20 = -20 + 40v
v = 40/40 = 1 m/s towards right
CASE - 4 => When A catches the ball , then applying L.C.L.M.
-4 x 5 + (-0.5) x 40 = -44 v
-20 - 20 = -44 v
v = 10/11 m/s towards left
(c) CASE - 5 => When A throws the ball , then applying L.C.L.M.
44 x (10/11) = 4 x 5 - 40 x v
v = 60/40
v = 3/2 m/s towards left
CASE - 6 => When B receives the ball , then applying L.C.L.M.
40 x 1 + 4 x 5 = 44 x v
v = 60/44
v = 15/11 m/s towards right
CASE - 7 => When B throws the ball , then applying L.C.L.M.
44 x (66/44) = -4 x 5 + 40 x v
v = 80/40
v = 2 m/s towards right
CASE - 8 => When A catches the ball , then applying L.C.L.M.
-4 x 5 -40 x (3/2) = -44 v
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 60/11 m/s > 5 m/s .
So , it can't catch the ball .
So , it can only roll the ball six.
(e) Let the ball and the body A at the initial position be at origin.
Centre of mass of the system =( 40 x 0 + 4 x 0 + 40 x d ) / (40 + 40 + 4 )
= 40d / 84
= 10d/11
Hope it helps!
Answer:
Explanation:(a) CASE - 1 => Total momentum of the man A and the ball will remain constant
Therefore ,
0 = 4 x 5 - 40 x v
v = 0.5 m/s towards left
(b) CASE - 2 => When B catches the ball , the momentum between the B and the ball will remain constant.
4 x 5 = 44v
v = 20/44 m/s
CASE - 3 => When B throws the ball , then applying Law of Conservation of Linear Momentum (L.C.L.M.)
44 x (20/44) = -4 x 5 + 40 x v
20 = -20 + 40v
v = 40/40 = 1 m/s towards right
CASE - 4 => When A catches the ball , then applying L.C.L.M.
-4 x 5 + (-0.5) x 40 = -44 v
-20 - 20 = -44 v
v = 10/11 m/s towards left
(c) CASE - 5 => When A throws the ball , then applying L.C.L.M.
44 x (10/11) = 4 x 5 - 40 x v
v = 60/40
v = 3/2 m/s towards left
CASE - 6 => When B receives the ball , then applying L.C.L.M.
40 x 1 + 4 x 5 = 44 x v
v = 60/44
v = 15/11 m/s towards right
CASE - 7 => When B throws the ball , then applying L.C.L.M.
44 x (66/44) = -4 x 5 + 40 x v
v = 80/40
v = 2 m/s towards right
CASE - 8 => When A catches the ball , then applying L.C.L.M.
-4 x 5 -40 x (3/2) = -44 v
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 60/11 m/s > 5 m/s .
So , it can't catch the ball .
So , it can only roll the ball six.
(e) Let the ball and the body A at the initial position be at origin.
Centre of mass of the system =( 40 x 0 + 4 x 0 + 40 x d ) / (40 + 40 + 4 )
= 40d / 84
=10d/21