Question

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Answer 1

Answer:

5m/swill be the right answer.

Explanation:

hope it helps you dude

Answer 2

SOLUTION.

You encountered such questions many a time.

Aid to these type of question is obtaining two equation by using law of conservation of momentum ( in usefull direction ). and second obtain from coefficient of restitution.

MOMENTUM CONSERVATION...

Initial velocity of 2m in X direction = v(1)

Final velocity of 2m in X direction = v(2) cos60

Initial velocity of m in X direction = 0

Final velocity of m in X direction = v(2) cos60

Where v(2) is the velocity of combine mass

MOMENTUM CONSERVATION IN X DIRECTION

2mv(1) + m(0) = (2m + m)v(2)cos60

2mv(1) = 3mv(2)1/2

2v(1) =( 3/2) v(2).......EQ(1)

APPLY MOMENTUM CONVERSATION IN Y DIRECTION

Initial velocity of m in y direction = 10√3

Final velocity of m in y direction = v(2) sin60

Initial velocity of 2m in y direction = 0

Final velocity of 2m in y direction = v(2)sin60

apply momentum conversation

m(10√3) + 0 = (m + 2m)v(2)sin60

10√3 = 3v(2)(√3/2)

from here we get.....

v(2) = 20/3

put this velocity in EQ(1)

2v(1) = 3/2 (20/3)

v(1) = 5....(Answer)