Considering a system having two masses `m_(1)` and `m_(2)` in which first mass is pushed towards centre of mass by a distance a, the distance required to be moved for second mass to keep centre of mass at same position is :- A. `(m_(1))/(m_(2))a`B. `(m_(1)m_(2))/(a)`C. `(m_(2))/(m_(1))a`D. `((m_(2)m_(1))/(m_(1) + m_(2)))a`

Considering a system having two masses `m_(1)` and `m_(2)` in which fi

1.	Considering a system having two masses `m_(1)` and `m_(2)` in which first mass is pushed towards centre of mass by a distance a, the distance required to be moved for second mass to keep centre of mass at same position is :- A. `(m_(1))/(m_(2))a`B. `(m_(1)m_(2))/(a)`C. `(m_(2))/(m_(1))a`D. `((m_(2)m_(1))/(m_(1) + m_(2)))a`
Answer» Correct Answer - A `Deltabar(x) = (m_(1).Deltax_(1) + m_(2).Deltax_(2))/(m_(1) + m_(2))` `rArr 0 = (m_(1)a + m_(2)Deltax_(2))/(m_(1) + m_(2))` `rArr Deltax_(2) = -(m_(1)a)/(m_(2))`

Discussion

No Comment Found

Related InterviewSolutions

Two blocks `A` and `B` are joined by means of a slacked string passing over a massless pulley as shown in Figure. The system is released from rest and it becomes taut when `B` falls a distance `0.5 m`. a. Find the common velocity of the two blocks just after the string becomes taut. b. Find the magnitude of impulse on the pulley by the clamp during the small interval while string becomes taut.
A body of mass `1 kg` moving with velocity `1 m//s` makes an elastic one dimesional collision with an identical stationary body. They are in contact for brief time `1 sec`. Their force of interaction increases form zero to `F_(0)` linearly in time `0.5s` and decreases linearly to zero in further time `0.5 sec` as shown in figure. Find the magnitude of force `F_(0)` in newton.
Consider a two particle system with particles having masses `m_1 and m_2` if the first particle is pushed towards the centre of mass through a distance d, by what distance should the second particle is moved, so as to keep the center of mass at the same position?A. dB. `(m_(2))/(m_(1))d`C. `(m_(1))/(m_(1)+m_(2))d`D. `(m_(1))/(m_(2))d`
Two paricle A and B initially at rest, move towards each other under mutual force of attraction. At the instant when the speed of A is V and the speed of B is 2V, the speed of the centre of mass of the system isA. `3`B. `v`C. `1.5v`D. zero
A body of mass `m = 3.513 kg` is moving along the x-axis with a speed of `5.00ms^(-1)`. The magnetude of its momentum is recorded asA. `17.565 kg ms^(-1)`B. `17.56 kgms^(-1)`C. `17.57 kg ms^(-1)`D. `17.6 kg ms^(-1)`
Two paricle A and B initially at rest, move towards each other under mutual force of attraction. At the instant when the speed of A is V and the speed of B is 2V, the speed of the centre of mass of the system isA. velocity remains constantB. zeroC. 2vD. 3v/2
Two paricle A and B initially at rest, move towards each other under mutual force of attraction. At the instant when the speed of A is V and the speed of B is 2V, the speed of the centre of mass of the system isA. zeroB. vC. 1.5vD. 3v
When two blocks connected by a spring move towards each other under mutual interactionA. their velocities are equal and opposite.B. their accelerations are unequal and oppositeC. the force acting on them are equal and oppositeD. their momentum are equal and opposite
A rocket, with an initial mass of `1000kg`, is launched vertically upwards from rest under gravity. The rocket burns fuel at the rate of `10kg` per second. The burnt matter is ejected vertically downwards with a speed of `2000ms^-1` relative to the rocket. If burning ceases after one minute, find the maximum velocity of the rocket. (Take g as constant at `10ms^-2`)
An initially stationary box on a frectionless floor explodes into two places, places A with mass `m_(A)` and piece B with mass `m_(B)`. Two pieces then move across the floor along x-axis. Graph of position versus time for the two pieces are given. If all the graphs are possible then, in which of the following cases external force must be acting on the box :-A. IIB. VC. VID. I

Discussion

No Comment Found

Related InterviewSolutions

Reply to Comment