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The blocks shown in the figure have equal masses. The surface of A is smooth but that of B has a friction coefficient of $0.10$ with the floor. Block A is moving at a speed of $10\,m{s^{ - 1}}$ towards B which is kept at rest. Find the distance travelled by B if $\left( {Take\,g = 10\,m{s^{ - 2}}} \right)$
a) the collision is perfectly elastic and
b) the collision is perfectly inelastic.

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Last updated date: 01st Mar 2024
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IVSAT 2024
Answer
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Hint: In the question, velocity of the first object and the friction coefficient of A and B is given. By substituting the values in the equation of the distance and relating the equation of mass and velocity, we get the value of the distance of the second object.
Formula used
The expression for finding the distance is,
$W = mg \times d$
Where,
$m$ be the mass, $g$ be the acceleration due to the gravity and $d$ be the distance.

Complete step by step solution
Given that
Velocity of the first object ${v_1} = 10\,m{s^{ - 1}}$
Velocity of the second object ${v_2} = 0$
${v_1}$ and ${v_2}$ be the velocity of $A$ and $B$ respectively after the collision

a) the collision is perfectly elastic
${m_1} + m{v_2} = {v_1} + {v_2}$
Substitute the known values in the above equation, we get
$\Rightarrow 10 + 0 - {v_1} + {v_2}$
Simplify the above equation we get,
$\Rightarrow {v_1} + {v_2} - 10.............\left( 1 \right)$
Now, ${v_1} - {v_2} = - \left( {{v_1} - {v_2}} \right)$
Substitute the known values in the above equation, we get
$\Rightarrow {v_1} - {v_2} = - \left( {10 - 0} \right)$
$\Rightarrow {v_1} - {v_2} = - 10............\left( 2 \right)$
Subtracting the equation $2$from the equation $1$, we get
$\Rightarrow 2{v_2} - 20$
Simplify the above equation we get,
$\Rightarrow {v_2} = 10\,m{s^{ - 1}}$
Now we take the deceleration of B, we get
$B = \mu g$
According to the work energy principle
$\Rightarrow 0.5 \times m \times {0^2} - \left( {0.5} \right) \times m \times {v_2}$
$\Rightarrow W = - \mu \times mg \times d$
Here $d$ is the distance travelled by $B$
$\Rightarrow d = \dfrac{{100}}{{2 \times 0.1 \times 10}}$
$\Rightarrow d = 50\,m$

b) the collision is perfectly inelastic.
$\Rightarrow {m_1} \times {u_1} + m \times {u_2} = \left( {m + m} \right) \times v$
Substitute the known values in the above equation, we get
$\Rightarrow {m_1} \times 10 + m \times 0 = \left( {m + m} \right) \times v$
$\Rightarrow {m_1} \times 10 + m \times 0 = \left( {2m} \right) \times v$
$\Rightarrow v = \dfrac{{10}}{2}$
$\Rightarrow v = 5\,m{s^{ - 1}}$
Now, the two blocks are moved together and sticking to each other, we get
Now apply the principle of work energy, we get
$\Rightarrow \left( {0.5} \right) \times 2\,m \times {0^2} - \left( {0.5} \right) \times 2\,m \times {v^2}$
$\Rightarrow {d_2} = {5^2}\left( {0.1 \times 10 \times 2} \right)$
$\Rightarrow {d_2} = 12.5\,m$

Therefore, the distance travelled by the $B$ is $12.5\,m$.

Note: In the question, we find the value of before collision and after collision and then we equate the values because the collision is perfectly elastic so we get the value of the distance of the second object. But in the perfectly inelastic we equate the value of the two masses then we find the value of the distance of the second object.