Question

A spring is compressed between two toy carts of masses $m_1$ and $m_2$. When the toy carts are released, then spring exerts on each, equal and opposite average force for the same time $t$. Assume that there is no friction between the toy carts and the ground, then the speeds of toy carts are:

• A. In opposite directions but in inverse ratio of masses
• B. In the opposite directions but in the direct ratio of masses
• C. Equal but in opposite direction
• D. Equal but in same direction

Answer 1

The correct option is A

In opposite directions but in inverse ratio of masses

Step 1: Given data

Mass of one carts $=m_1$

Mass of second carts $=m_2$

Let Acceleration of one carts$=a_1$

Let Acceleration of second carts$=a_2$

Let Velocity of one carts$=v_1$

Let Velocity of second carts$=v_2$

Step 2: Find the speeds of toy carts

Let forces exerted by carts are $F_1$and $F_2$

By using Newton's second law,

$$\begin{gathered} F=ma \\ {F}_1=m_1{a}_1 \\ {F}_2=m_2{a}_2 \end{gathered}$$

Since, force is equal and opposite. Then,

$$\begin{gathered} F_1=-F_2 \\ {m}_1{a}_1=-m_2{a}_2 \end{gathered}$$

As, the time is same. Using acceleration as

$$\begin{gathered} a_1=\frac{v_1}{t} \\ {a}_2=\frac{v_2}{t} \\ \frac{m_1{v}_1}{t}=-\frac{m_2{v}_2}{t} \\ \frac{v_1}{v_2}=-\frac{m_2}{m_1} \end{gathered}$$

Thus, the speeds of toy carts is in opposite directions but in inverse ratio of masses.

Hence, option A is correct.