You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XII

Physics

Law of Conservation of Mechanical Energy

Two bodies of...

Question

Two bodies of masses $3 k g$ and $2 k g$ are connected by a long string and the string is made to pass over a smooth fixed pulley. Initially the bodies are held at the same level and released from rest. The velocity of the $3 k g$ body after one second is
(take $g = 10 m / s^{2}$ )

2 m/s

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

1 m/s

No worries! We‘ve got your back. Try BYJU‘S free classes today!

0.4 m/s

No worries! We‘ve got your back. Try BYJU‘S free classes today!

4 m/s

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is A 2 m/s
If a is the acceleration of the system, then the equation of motion of the system,
$(m_{1} + m_{2}) a = F_{n e t} = (m_{1} - m_{2}) g$
$a = \frac{m_{1} - m_{2}}{m_{1} + m_{2}} g = 2 m / s^{2}$
Now using , $v = u + a t$ where u=0 and t=1 s
$v = a t = 2 \times 1 = 2 m / s$

Suggest Corrections

Similar questions

Q. Two blocks of masses

3 k g

and

2 k g

respectively are tied to the ends of a string which passes over a light frictionless pulley. The masses are held at rest at the same horizontal level and then released. The distance traversed by centre of mass in

5

seconds is :

(g = 10 m / s^{2})

Q. Two masses, (M+m) and (M-m) are attached to the ends of a light inextensible string and the string is made to pass over the surface of a smooth fixed pulley. When the masses are released from rest, the acceleration of the system is

Q. Two blocks of masses

3 kg

and

2 kg

respectively are tied to the ends of a string which passes over a light pulley as shown in the figure below. The masses are held at rest at the same horizontal level and then released. The distance moved by

C O M

5

seconds is

Two bodies of mass 3 kg and 4 kg are suspended at the ends of massless string passing over a frictionless pulley. The acceleration of the system is (g=9.8 $m / s^{2}$ )

Q. Two unequal masses

m_{1}

and

m_{2}

are connected by a string going over a clamped light smooth pulley as shown in figure

m_{1} = 3 k g

and

m_{2} = 6 k g

. The system is released from rest (a) Find the distance traveled by the first block the first two seconds.

g = 10 m / s^{2}

Join BYJU'S Learning Program

Two bodies of masses 3kg and 2kg are connected by a long string and the string is made to pass over a smooth fixed pulley. Initially the bodies are held at the same level and released from rest. The velocity of the 3kg body after one second is (take g=10m/s2)

The correct option is A 2 m/sIf a is the acceleration of the system, then the equation of motion of the system,(m1+m2)a=Fnet=(m1−m2)ga=m1−m2m1+m2g=2m/s2Now using , v=u+at where u=0 and t=1 sv=at=2×1=2m/s

Two bodies of masses $3 k g$ and $2 k g$ are connected by a long string and the string is made to pass over a smooth fixed pulley. Initially the bodies are held at the same level and released from rest. The velocity of the $3 k g$ body after one second is
(take $g = 10 m / s^{2}$ )