Question 33: Laws of Motion

Question

The masses of block $$A$$ and $$B$$ are $$m$$ and $$M$$, respectively. Between $$A$$ and $$B$$, there is a constant frictional force $$F$$ and $$B$$ can slide on a smooth horizontal surface. A is set in motion with velocity $$3 v_0$$ while $$B$$ is at rest. What is the distance moved by $$A$$ relative to $$B$$ before they move with the same velocity?

Accepted Answer

Consider the FBD as given below.
Equations of motion,
$$\begin{aligned}
& a_A=\frac{F}{m}(	ext { in }-x 	ext { direction }) \
& a_B=\frac{F}{M}(	ext { in }+x 	ext { direction })
\end{aligned}$$
Thus, we have, relative accleration of $$A$$ w.r.t. $$B$$ as
$$\begin{aligned}
a_{A B} & =a_A-a_B=\frac{-F}{m}-\frac{F}{M} \
& =-F\left(\frac{M+m}{M m}ight) \quad 	ext{(along} ~\mathrm{-x}-	ext{direction)}
\end{aligned}$$
Initial relative velocities of $$A$$ w.r.t $$B, u_{A B}=3 v_0$$ using equation $$v^2-u^2+2 a s$$,
$$\begin{aligned}
0 & =9 v_0^2-\frac{2 F(M+m)}{M m} s \
\Rightarrow \quad s & =\frac{9 M m v_0^2}{2 F(M+m)}
\end{aligned}$$
Distance moved relative to $$B, S_{A B}=\frac{9 M m v_0^2}{2 F(M+m)}$$

Answer

$$\frac{m M v_0^2}{F(m-M)}$$

Answer

$$\frac{M v_0^2}{18 F(m-M)}$$

Answer

$$\frac{m M v_0^2}{9 F(m+M)}$$

Answer

$$\frac{9 m M v_0^2}{2 F(m+M)}$$

Laws of Motion

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