You pull a block of mass macross a frictionless table with a constant force. You also pull with an equal constant force a block of larger mass M. The blocks are initially at rest. If you pull the blocks through the same distance, which block has the greater kinetic energy, and which block has the greater momentum? If instead you pull the blocks for the same amount of time, which block has the greater kinetic energy, and which block has the greater momentum?

The given data is listed below as follows,

• The mass of the smaller block is,$m$
• The mass of the larger block is,$M$

The work done is described as the product of the force exerted on an object and the distance moved by the object.

Newton’s first law states that a body will remain at rest unless an external force acts it.

The equation of the work done is expressed as:

$W=F.d$ …(i)

Here, $F$is the force acting and $d$is the distance moved by the object.

As both the blocks are at rest, the work done is zero, which is the same for both the small and the large blocks.

According to the energy principle, the amount of the work done is converted to kinetic energy, and the blocks will have the same amount of kinetic energy as they have the same amount of work done.

The equation of the kinetic energy of the blocks is expressed as:

$$\begin{gathered} k=\frac{p^2}{2m} \\ p=\sqrt{2km} \end{gathered}$$ …(ii)

Here,$p$ is the momentum and the mass of the object.

From the above expression, it can be observed that the mass is directly proportional to the momentum; then, if the mass of the object increases, the momentum will also increase.

Hence, the block with mass$M$ has greater momentum than the block with mass$m$ .

From the second equation of motion, the relationship between the distance and the time is expressed as follows,

$d=ut+\frac{1}{2}a{t}^2$

Here, $d$is the distance, $u$is the initial speed, $t$is the time, and $a$is the acceleration that is $a=\frac{F}{m}$.

As both the blocks were at rest, the initial velocity would be zero. Hence, the above equation can be written as follows:

$$\begin{gathered} d=\frac{1}{2}a{t}^2 \\ =\frac{1}{2}\frac{F}{m}{t}^2 \end{gathered}$$

Substitute the above values in the equation (i),

role="math" localid="1657084414073" $$\begin{gathered} W=F.\frac{1}{2}\frac{F}{m}{t}^2 \\ =\frac{F^2{t}^2}{2m}\dots \left(iii\right) \end{gathered}$$

From the above expression, it can be observed that the mass is inversely proportional to the work done. So, when mass is increased, then the kinetic energy will decrease.

According to the energy principle, the work done is converted to kinetic energy. So, the mass with a smaller value will have greater kinetic energy than the mass $M$.

Equate equation (i) and equation (ii) as the work done equals the kinetic energy.

$$\begin{gathered} \frac{F^2{t}^2}{2m}=\frac{p^2}{2m} \\ p=\sqrt{F.t} \end{gathered}$$

As both the blocks have the same applied force and timeframe, both will have the same amount of momentum.

Thus, in the first case, if the blocks are pulled through the same distance,the mass$M$ has greater momentum than the mass$m$ but the kinetic energy will be the same for both blocks.

And in the second case, if the blocks are pulled for the same amount of time, then the mass$m$ will have greater kinetic energy than the mass$M$ and the momentum of both the blocks will be the same.