This problem aims to familiarize us with a **friction-less motion** between two** masses **as a **single system.** The concept required to solve this problem includes **acceleration**, **newtons law of motion**, and the law of **conservation of momentum.**

In this particular problem, we require the help of **newton’s second law,** which is a** quantitative** definition of the **transformations** that a force can have on the **motion of a body**. In other words, it is the rate of change of the** momentum** of a body. This momentum of a body is equivalent to **mass** times its **velocity.**

For a body having a constant mass $m$, **Newton’s second law** can be composed in the form $F = ma$. If there are multiple **forces** acting on the body, it is equally **accelerated** by the equation. Contrarily, if a body doesn’t **accelerate,** no kind of **force** is acting on it.

## Expert Answer

The **force** $F = 250 \space N$ is causing **acceleration** to both the boxes.

Applying **Newton’s** second law to obtain the **acceleration** of the whole system:

\[ F = (m_A+ m_B)a_x\]

Making $a_x$ the subject of the equation.

\[ a_x = \dfrac{F}{(m_A+m_B)} \]

\[a_x = \dfrac{(250)}{20+5}\]

\[ a_x = 10 \space m/s^2 \]

As Box A is exerting **force** on Box B, both the boxes are **accelerating** at the same speed. So it can be said the **acceleration** of the whole system is $10\space m/s^2$.

Now applying the** Newtons second law** on box B and calculating the **force** $F$:

\[F_A = m_ba_x\]

\[= 5 \times 10\]

\[F_A = 50 \space N\]

## Numerical Answer:

Box A exerts the **force** of **magnitude** $50 \space N$ on box B.

## Example

Boxes A and B and C are in contact on a horizontal, **frictionless surface.** Box A has **mass** $20.0 kg$, box B has **mass** $5.0 kg$ and box C has a **mass** $15.0 kg$. A **horizontal force** of $200 N$ is exerted on box A. What is the **magnitude** of the **force** that box B exerts on box C and the Box A exerts on Box B?

The force $F = 200\space N$ is causing **acceleration** to all the boxes.

Applying **Newton’s second** law to acquire the acceleration of the entire system:

\[F = (m_A+m_B+m_C) a_x\]

Making $a_x$ the subject of the equation.

\[ a_x = \dfrac{F}{(m_A+m_B+m_C)} \]

\[ a_x = \dfrac{(200)}{20 +5+15} \]

\[ a_x = 5\space m/s^2\]

As Box A is exerting force on Box B, and then Box B is exerting force on Box C, All the boxes are **accelerating** at the same speed. So it can be said the **acceleration** of the whole system is $5\space m/s^2$.

Now applying the **Newtons second** law on box C and calculating the force $F_B$.

\[ F_B = m_Ca_x \]

\[= 15 \times 5\]

\[F_B = 75 \space N\]

Box B exerts the **force** of $75 \space N$ on Box C.

Now,

\[F_A = m_Ba_x\]

\[= 5 \times 5\]

\[F_A = 25 \space N\]

Box A exerts the **force** of $25 \space N$ on Box B.