Question

Guys....
Answer this

Two masses M and m are tied with a string and arranged

as shown.
Mass m is released from rest, assume that its
height above ground is greater than 2h initially.
The
velocity of block M just after it loses contact with ground is​

Answer 1

ANSWER:

option c is right.

EXPLANATION:

FOR MASS, m:

• tension acts towards the pulley in upward direction.
• It's weight mg acts downwards.
• Since the mass, m is dropped ,it is accelerated downwards.
• So the pseudo force Fp = ma acts upwards .

T = mg - ma

FOR MASS, M:

• tension acts upwards towards the pulley.
• Due to release of mass m ,the mass M is accelerated upwards.
• So the pseudo force, Fp = Ma acts downwards.

T = Ma

NOTE:

• Tension is same for both the masses.

• On equating tension we get acceleration ,which is also same for both.

• The mass m ,starts from rest and gains velocity , u.

• Total height it travelled = u² / 2g

• here the total height becomes 2h.

• Time for its travel , t = u / g

• the mass M gains velocity ,v for time t.

• It's initial velocity is zero ,since it starts from rest.

• For the same time t ,the mass M gains velocity.

• so the velocity of mass M can be found by applying first equation of motion, v = u + at.

SOLUTION:

refer to the attachment

Explanation: