You visited us 2 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XII

Physics

Motion Under Gravity

A mass of 3...

Question

A mass of $3 k g$ descending vertically downward support a mass of $2 k g$ by means of a light string passing over a pulley. At the end of $5 s$ the string breaks. How much high from now the $2 k g$ mass will go ? ( $g = 9.8 m / s^{2}$ )

4.9 m

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

9.8 m

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

16.9 m

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

2.45 m

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

Open in App

Solution

The correct option is A $4.9 m$
Given that,

Mass $m_{1} = 3 k g$

Mass $m_{2} = 2 k g$

Now, the initial acceleration a of the system

$a = \frac{3 g - 2 g}{3 + 2}$

$a = \frac{g}{5}$

At $5 s e c$

Now, from equation of motion

$v = u + a t$

$v = 0 + \frac{g}{5} \times 5$

$v = 9.8 m / s$

When the string breaks at $t = 5 s$ ,

The mass $2 k g$ more under gravity

So, height is

$h = \frac{v^{2}}{2 g}$

$h = \frac{{(9.8)}^{2}}{2 \times 9.8}$

$h = 4.9 m$

Hence, the height is $4.9 m$

Suggest Corrections

Similar questions

Q. A mass of

3 k g

descending vertically downward supports a mass of

2 k g

by means the end of

5 s

, the string breaks. How much higher the

2 k g

mass will go further?

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 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

2 k g

and

3 k g

are attached to the ends of the string passing over a pulley which is fixed at the top. The tension and acceleration in the string in terms of acceleration due to gravity,

g

are:

Q. A string with one end fixed on a rigid wall, passing over a fixed frictionless pulley at a distance of 2 m from the wall, has a point mass

M = 2 kg

attached to it at a distance of 1 m from the wall. A mass

m = 0.5 kg

attached at the free end is held at rest so that the string is horizontal between the wall and the pulley and vertical beyond the pulley. What will be the speed in m/s upto 1 decimal place with which the mass

M

will hit the wall when the mass m is released and

g = 9.8 {ms}^{- 2}

Join BYJU'S Learning Program

A mass of 3 kg descending vertically downward support a mass of 2 kg by means of a light string passing over a pulley. At the end of 5 s the string breaks. How much high from now the 2 kg mass will go ? (g=9.8 m/s2)

A mass of $3 k g$ descending vertically downward support a mass of $2 k g$ by means of a light string passing over a pulley. At the end of $5 s$ the string breaks. How much high from now the $2 k g$ mass will go ? ( $g = 9.8 m / s^{2}$ )