Question:

A spring is compressed between two toy carts of mass $m_1$ and $m_2$. When the toy carts are released, the springs exert equal and opposite average forces for the same time on each toy cart. If $V_1$ and $V_2$ are the velocities of the toy carts and there is no friction between the toy carts and the ground, then :

Updated On: Aug 16, 2024
  • $V_2 / V_2 = m_1 / m_2$
  • $V_1 / V_1 = m_2 / m_1$
  • $V_1 / V_2 = - m_2 / m_1$
  • $V_1 / V_2 = m_1 / m_2$
Hide Solution
collegedunia
Verified By Collegedunia

The Correct Option is C

Solution and Explanation

Here, $m_{1} \frac{d V_{1}}{d t}=-m_{2} \frac{d V_{2}}{d t}$ or $m_{1} a_{1}=m_{2} a_{2} \ldots$ (1) now, $V_{1}=a_{1} t$ and $V_{2}=a_{2} t$ (by using $v=u+$ at) from (1), $m _{1} \frac{ V _{1}}{ t }=- m _{2} \frac{ V _{2}}{ t }$ $\therefore \frac{ V _{1}}{ V _{2}}=-\frac{ m _{2}}{ m _{1}}$
Was this answer helpful?
1
0

Concepts Used:

Work

Work is the product of the component of the force in the direction of the displacement and the magnitude of this displacement.

Work Formula:

W = Force × Distance

Where,

Work (W) is equal to the force (f) time the distance.

Work Equations:

W = F d Cos θ

Where,

 W = Amount of work, F = Vector of force, D = Magnitude of displacement, and θ = Angle between the vector of force and vector of displacement.

Unit of Work:

The SI unit for the work is the joule (J), and it is defined as the work done by a force of 1 Newton in moving an object for a distance of one unit meter in the direction of the force.

Work formula is used to measure the amount of work done, force, or displacement in any maths or real-life problem. It is written as in Newton meter or Nm.