Question

No spam !!!

Correct answers accepted!!!

Content quality required!!!

Thanks!!​

Answer 1

Answer: 4.9 m

Explanation:

Let the resultant force acting downwards in the direction  of 3kg and acting upward in side of 2kg  mass be F and acceleration of masses be a.

Since , 3 kg is heavier than 2kg. So 3 kg mass will pull 2kg mass in upward direction.

Force acting downward by  3kg mass = 3g(m₁)

Force acting downward by 2kg mass  = 2g(m₂)

Resultant force(F) = 3g - 2g

F = g

Acceleration (a) = F/(m₁+m₂)

a = g/(3+2)

a = g/5 m/s²

Case I:- Firstly m₂ starts moving this acceleration 'a' in upward direction till string does not get break i.e given as 5 seconds

So in this case we have,

Initial velocity (u) = 0

Final velocity (v) = ?

time(t) = 5 sec

So by using first equation of motion

v = u + at

v = 0 + g/5 × 5

v = g m/s

Case II :- Now m₂ is moving with this acceleration 'a' in upward direction. At the moment when that string breaks, m₂ will still  move in upward direction with initial speed as 'g' m/s as per Law of Inertia and its velocity at highest point will  be zero.

So in this case we have,

Initial velocity (u) = g

Final velocity (v) = 0

v² = u² - 2as

0 = g² - 2as

2as = g²

2gs = g²

2s = g

s = g/2

s = 9.8/2

s = 4.9 m

Answer 2

$\huge\sf\orange{Answer:}$

Option - A ( 4.9 m )

$\sf\orange{Explanation:}$

Let the acting force be 3kg for downwards.

Let the acting force be 2kg for upwards.

Let F be the mass and A be acceleation.

3kg > 2kg ( 3kg is greater than or more than 2kg hence, 3 kg mass taken to the 2kg mass for upward.

1) 3kg mass = 3g (m1)

2) 2kg mass = 2g (m2)

F = 3g - 2g

F = gA = F / ( m1 + m2)

A = g / ( 3 + 2 )

A = g / 5 ( m/s² )

Case 1

u = 0

v = To find (?)

v = g ( m/s )

Case 2

u = g

v = 0

v² = u² = 2as

0 = g²

2as = g²

2gs = g²

2s = g

s = g / 2

s = 9.8 / 2

.°. s = 4.9 m

Therefore, the height from now the 2 kg mass will go to ( 4.9 m ).