Question

A strip wood of mass \[M\] and length \[l\] is placed on a smooth horizontal surface. An insect of mass \[m\]starts at one end of the strip and walks to the other end in time \[t\], moving with a constant speed. The speed of the insect as seen from the ground is:
 \[\begin{align}
  & A.\dfrac{l}{t}\left( \dfrac{M}{M+m} \right) \\
 & B.\dfrac{l}{t}\left( \dfrac{m}{M+m} \right) \\
 & C.\dfrac{l}{t}\left( \dfrac{M}{m} \right) \\
 & D.\dfrac{l}{t}\left( \dfrac{m}{M} \right) \\
\end{align}\]

Answer 1

Hint: We will draw a figure to understand the motion clearly. Then we will use the law of conservation of momentum to obtain relation in motion between the insect and the strip. We must be very aware of the frame of reference and we will find the velocity of the insect from the equation of relative velocity because we need velocity on the ground frame.

Formula used:
\[{{m}_{1}}{{u}_{1}}+{{m}_{2}}{{u}_{2}}={{m}_{1}}{{v}_{1}}+{{m}_{2}}{{v}_{2}}\]

Complete step by step answer:
Firstly let’s draw the figure to understand the situation more clearly.
At the initial condition, both the strip and the insect is at rest. So the initial momentum will be zero and from the momentum conservation formula, we can obtain the relation,

\[\begin{align}
  & mv-MV=0 \\
 & \Rightarrow mv=MV \\
\end{align}\]

Where, \[m\] and \[v\] is mass and velocity of insect respectively on the strip frame and, \[M\] and \[V\] are the mass and velocity of the strip respectively from the strip frame.
When we take reference of the ground frame the velocity of insect will be relative to the strip which is given by,
\[{{v}_{is}}=v-\left( -V \right)=v+V\]
Now we will use the basic velocity equation for the insect, there it travels the distance \[l\] on the strip with time taken\[t\].
\[\Rightarrow v+V=\dfrac{l}{t}\]
That means the velocity of insect with ground frame is given as,

\[\begin{align}
  & {{v}_{g}}=v+V=\dfrac{l}{t} \\
 & \Rightarrow {{v}_{g}}<\dfrac{l}{t} \\
\end{align}\]

But there is no such answer given in the options. So we will carry the question further to find the velocity of the strip from the ground frame.
From the momentum conservation equation, we have \[mv=MV\]
If we substitute the velocity of insect in this expression with the velocity of insect from ground frame, we will get the velocity of the strip from the ground frame.
    i.e.
\[\begin{align}
  & mv=MV \\
 & \Rightarrow v=\dfrac{MV}{m} \\
\end{align}\]

Now we will substitute this velocity in the equation for velocity of the insect on the ground frame.

\[\begin{align}
  & v+V=\dfrac{l}{t} \\
 & \Rightarrow \left( \dfrac{MV}{m} \right)+V=\dfrac{l}{t} \\
\end{align}\]
\[\Rightarrow \left( \dfrac{M}{m}+1 \right)V=\dfrac{l}{t}\]
So, velocity of strip from ground frame is \[V=\dfrac{l}{t}\left( \dfrac{m}{M+m} \right)\]
Now, this is present in the options given. So we can select the right answer as option b.

Note:
There is a problem regarding the options given and the question here. So we have extended the question to find the velocity of the strip from the ground frame. We must be very careful while assigning velocity terms because they may vary from frame to frame. So there is a chance for mistakes happening in finding the right choice.