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

Newton's Laws for System of Particles

The system of...

Question

The system of two weight with masses $m_{1}$ and $m_{2}$ is connected with a weightless spring as shown. The system is resting on support $S$ . The support $S$ is quickly removed. The acceleration of each of the weights right after the support $S$ is removed is ( $a_{1}$ is acceleration of $m_{1}$ and $a_{2}$ that of $m_{2}$ )

a_{1} = 0, a_{2} = \frac{(m_{1} + m_{2}) g}{m_{2}}

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

a_{1} = 0, a_{2} = \frac{(m_{1} + m_{2}) g}{m_{1}}

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

a_{1} = \frac{(m_{1} + m_{2}) g}{m_{1}}, a_{2} = 0

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

a_{1} = 0,

a_{2} = 0

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

Open in App

Solution

The correct option is A $a_{1} = 0, a_{2} = \frac{(m_{1} + m_{2}) g}{m_{2}}$
When the support is removed, spring force does not change abruptly.

Old $F B D$
the system is in equilibrium
For $m_{1} : m_{1} g - k x = 0$
$\Rightarrow k x = m_{1} g$

New $F B D$ .
For $m_{1} : m_{1} g - k x = m_{1} a_{1}$
$m_{1} g - m_{1} g = m_{1} a_{1}$
$\Rightarrow a_{1} = 0$
For $m_{2} : m_{2} g + k x = m_{2} a_{2}$
$m_{2} g + m_{1} g = m_{2} a_{2}$
$\Rightarrow a_{2} = \frac{(m_{1} + m_{2}) g}{m_{2}}$

Suggest Corrections

Similar questions

The system of two weights 1 & 2 with masses $M_{1}$ and $M_{2}$ are connected with weightless spring as shown. The system is resting on the support 'S'. Find the acceleration of each of the weights just after the support 'S' is quickly removed.

Q. The centre of mass of a system of two particles of masses

m_{1}

and

m_{2}

is at a distance

a_{1}

from mass

m_{1}

and at a distance

a_{2}

from mass

m_{2}

such that

Q. The system of two weights with masses

m_{1}

and

m_{2}

are connected with weightless spring as shown in

F i g .6 .211

The system is resting on the support S. Find the acceleration of each of the wights just after the support S is quickly removed.

Q. The acceleration of '

m_{1}

' and '

m_{2}

' are '

a_{1}

' and '

a_{2}

' then:

In the shown diagram m $_{1}$ = m $_{2}$ = 4 kg and m $_{3}$ = 2 kg.Coefficient of friction between m $_{1}$ and m $_{2}$ is 0.5. The mass m $_{1}$ is given a velocity v and it just stops at the other end of the mass m $_{2}$ in 2 sec. Let a $_{1}$ , a $_{2}$ and a $_{3}$ be the acceleration m $_{1}$ , m $_{2}$ and m $_{3}$ respectively, then:

Join BYJU'S Learning Program

The system of two weight with masses m1 and m2 is connected with a weightless spring as shown. The system is resting on support S. The support S is quickly removed. The acceleration of each of the weights right after the support S is removed is (a1 is acceleration of m1 and a2 that of m2 )

The correct option is A a1=0,a2=(m1+m2)gm2When the support is removed, spring force does not change abruptly.Old FBDthe system is in equilibriumForm1:m1g−kx=0⇒kx=m1gNew FBD.For m1:m1g−kx=m1a1m1g−m1g=m1a1⇒a1=0For m2:m2g+kx=m2a2m2g+m1g=m2a2⇒a2=(m1+m2)gm2