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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?


ANSWER:

(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

Comments

  1. UnknownAugust 6, 2019 at 3:13 AM

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  2. UnknownAugust 12, 2019 at 12:40 PM

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  3. UnknownAugust 18, 2019 at 9:01 AM

    https://www.chegg.com/homework-help/questions-and-answers/40-kg-homogeneous-plate-suspended-lour-wires-determine-tension-wire-fig-p518-q3194652

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  4. UnknownAugust 25, 2019 at 7:11 AM

    Help with this please
    https://www.chegg.com/homework-help/questions-and-answers/evaluate-integralgamma-z-ne-1-z-dz-gamma-circle-radius-1-centered-0-traveled-counterclockw-q20451639

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  5. AlexandraNovember 16, 2019 at 8:09 AM

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