Question

Figure shows two small blocks placed on a smooth horizontal surface. They start moving from the same line with force $2F$ and $F$ respectively acting on the blocks. Their momenta and kinetic energies at the instants of crossing the finishing line as shown in figure are $p_A,p_B$ and $K_A,K_B$. Then choose the correct option.

• A. $K_A=K_B,p_A=p_B$
• B. $K_A>K_B,p_A>p_B$
• C. $K_A>K_B,p_A<p_B$
• D. $K_A>K_B,p_A=p_B$

Answer 1

The correct option is D $K_A>K_B,p_A=p_B$

Let the displacement of the two blocks on the smooth horizontal surface be '$S$'.

Work done in moving block $A$, $\left(W_A\right)=2FS$

Work done in moving block $B$, $\left(W_B\right)=FS$

From work-energy theorem,
Work done = change in kinetic energy
Since the blocks are initially at rest, we can write that,
$\frac{W_A}{W_B}=\frac{K_A}{K_B}$

$\Rightarrow \frac{K_A}{K_B}=\frac{2FS}{FS}=\frac{2}{1}...\left(1\right)$

Hence, $K_A>K_B$

But, kinetic energy $\left(K\right)=\frac{p^2}{2m}$
$\Rightarrow K_A=\frac{p_A^2}{2m_A}and{K}_B=\frac{p_B^2}{2m_B}$
From this ,
$\frac{K_A}{K_B}=\frac{p_A^2}{p_B^2}\times \frac{m_B}{m_A}$
Using $\left(1\right)$ and the data given in the question

$\frac{2}{1}=\frac{p_A^2}{p_B^2}\times \frac{2m}{m}\Rightarrow p_A=p_B$

Thus, option (d) is the correct answer.