Answer
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Hint:Applying the formula of final velocities after a collision of two bodies of unequal mass we will find a relation between the velocities of stone and truck. Upon substitution of appropriate values we will be able to determine the velocity with which the stone flies away.
Formulae used: Final velocity of heavy particle (truck) after collision: ${v_1} = {u_1}$
Where ${v_1}$ is the final velocity of the truck and Is expressed in meter per second $(m{s^{ - 1}})$ and ${u_1}$ initial velocity of the truck and Is expressed in meter per second $(m{s^{ - 1}})$.
Final velocity of lighter particle (stone) after collision:: ${v_2} = 2{u_1} - {u_2}$
Where ${v_2}$ is the final velocity of the stone and Is expressed in meter per second $(m{s^{ - 1}})$ and ${u_2}$ initial velocity of the stone and Is expressed in meter per second $(m{s^{ - 1}})$.
Step by step solution:
From the question we gather the following information,
Final velocity of the stone which will come at rest after the collision $ = {u_2} = 0m{s^{ - 1}}$
Initial velocity of the truck $ = {u_1} = 10m{s^{ - 1}}$
We know that during an elastic between two objects of unequal weight, the higher body has a final velocity equal to the initial velocity of the lighter object. Mathematically we write it as ${v_1} = {u_1}$.
This is so because due to the huge difference in their masses, that the effect of collision on the heavier object is inconsequential.
But for the lighter object, the final velocity will be dependent on the initial velocity of the heavier object and its own. Mathematically we write it as ${v_2} = 2{u_1} - {u_2}$.
Substituting the values we find out the final velocity of the stone.
Therefore,
$
{v_2} = 2{u_1} - {u_2} = (2 \times 10) - 0 \\
\Rightarrow {v_2} = 20m{s^{ - 1}} \\
$
In conclusion, the correct option is B.
Note: Normally calculating the velocities, both initial and final, will give incorrect results. It is important to keep in mind the negligible effect the velocity of the lighter object has on the heavier object. The collision is not perfectly elastic and so we need to consider the different velocities.
Formulae used: Final velocity of heavy particle (truck) after collision: ${v_1} = {u_1}$
Where ${v_1}$ is the final velocity of the truck and Is expressed in meter per second $(m{s^{ - 1}})$ and ${u_1}$ initial velocity of the truck and Is expressed in meter per second $(m{s^{ - 1}})$.
Final velocity of lighter particle (stone) after collision:: ${v_2} = 2{u_1} - {u_2}$
Where ${v_2}$ is the final velocity of the stone and Is expressed in meter per second $(m{s^{ - 1}})$ and ${u_2}$ initial velocity of the stone and Is expressed in meter per second $(m{s^{ - 1}})$.
Step by step solution:
From the question we gather the following information,
Final velocity of the stone which will come at rest after the collision $ = {u_2} = 0m{s^{ - 1}}$
Initial velocity of the truck $ = {u_1} = 10m{s^{ - 1}}$
We know that during an elastic between two objects of unequal weight, the higher body has a final velocity equal to the initial velocity of the lighter object. Mathematically we write it as ${v_1} = {u_1}$.
This is so because due to the huge difference in their masses, that the effect of collision on the heavier object is inconsequential.
But for the lighter object, the final velocity will be dependent on the initial velocity of the heavier object and its own. Mathematically we write it as ${v_2} = 2{u_1} - {u_2}$.
Substituting the values we find out the final velocity of the stone.
Therefore,
$
{v_2} = 2{u_1} - {u_2} = (2 \times 10) - 0 \\
\Rightarrow {v_2} = 20m{s^{ - 1}} \\
$
In conclusion, the correct option is B.
Note: Normally calculating the velocities, both initial and final, will give incorrect results. It is important to keep in mind the negligible effect the velocity of the lighter object has on the heavier object. The collision is not perfectly elastic and so we need to consider the different velocities.
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