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Two steel balls $A$ and $B$ of mass $10kg$ and $10g$ roll towards each other with $5m/s$ and $1m/s$ on a smooth floor. After the collision with what speed $B$ moves if it is the case of an elastic collision?
$\left( a \right){\text{ 8m/s}}$
$\left( b \right){\text{ 10m/s}}$
$\left( c \right){\text{ 11m/s}}$
$\left( d \right){\text{ Zero}}$

Last updated date: 13th Jun 2024
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Hint So for this question, we have mass and the velocity of each ball. So we will calculate the velocity by using the formula which will use all the values given. The formula is given below. Since momentum is conserved so the elastic collision will be equal to one.
Formula used:
Velocity for the elastic collision,
$V = \left( {\dfrac{{{m_2} - e{m_1}}}{{{m_1} + {m_2}}}} \right){u_2} + \left( {\dfrac{{{m_1} - e{m_1}}}{{{m_1} + {m_2}}}} \right){u_1}$
$V$, will be the velocity
$e$, will be the elastic collision
${m_1}$, will be the mass of the ball $A$
${m_2}$, will be the mass of the ball $B$
${u_2}$, will be the initial velocity of the ball $A$
${u_1}$, will be the initial velocity of the ball $B$

Complete Step By Step Solution
Firstly, we will see the values given to us.
${u_1} = 5m/s$
${u_2} = 1m/s$
${m_1} = 10kg$
${m_2} = 10g = 0.01kg$
Now by using the formula we have already seen,
$V = \left( {\dfrac{{{m_2} - e{m_1}}}{{{m_1} + {m_2}}}} \right){u_2} + \left( {\dfrac{{{m_1} - e{m_1}}}{{{m_1} + {m_2}}}} \right){u_1}$
Velocity will be,
Substituting the values, we get
Here, the value of elastic collision will be
$ \Rightarrow e = 1$
On putting the values, we get
$ \Rightarrow V = \left( {\dfrac{{0.01 - 10}}{{10 + 0.01}}} \right)\left( { - 1} \right) + \left( {\dfrac{{10 - 10}}{{10 + 0.01}}} \right)5$
On further simplifying it, we get
\[ \Rightarrow V = \dfrac{{9.99 + 150}}{{10.01}}\]
Again solving the equation,
\[ \Rightarrow V = \dfrac{{109.99}}{{10.01}}\]
And it will be approximately equal to
$ \Rightarrow \approx 11m/s$
So if it is the case of elastic collision $B$ will move with the speed of approximately $11m/s$.

Hence the option $\left( c \right)$ is the correct option.

Note All collisions, from elastic to completely inelastic and anything in between, must conserve momentum. The reason is simply that all forces in a collision are internal to the objects colliding, i.e. no outside forces act on the system.
This is most effectively perceived in a two-body crash. The adjustment in the energy of an article is equivalent to the motivation, Assuming a consistent power, this is only the power times the timeframe. All the more, by and large, it is the fundamental of power concerning time.