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Question: Two steel balls \(A\) and \(B\) of mass \(10kg\) and \(10g\) roll towards each other with \(5m/s\) a...

Two steel balls AA and BB of mass 10kg10kg and 10g10g roll towards each other with 5m/s5m/s and 1m/s1m/s on a smooth floor. After the collision with what speed BB moves if it is the case of an elastic collision?
(a) 8m/s\left( a \right){\text{ 8m/s}}
(b) 10m/s\left( b \right){\text{ 10m/s}}
(c) 11m/s\left( c \right){\text{ 11m/s}}
(d) Zero\left( d \right){\text{ Zero}}

Explanation

Solution

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=(m2em1m1+m2)u2+(m1em1m1+m2)u1V = \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}
Here,
VV, will be the velocity
ee, will be the elastic collision
m1{m_1}, will be the mass of the ball AA
m2{m_2}, will be the mass of the ball BB
u2{u_2}, will be the initial velocity of the ball AA
u1{u_1}, will be the initial velocity of the ball BB

Complete Step By Step Solution
Firstly, we will see the values given to us.
u1=5m/s{u_1} = 5m/s
u2=1m/s{u_2} = 1m/s
m1=10kg{m_1} = 10kg
m2=10g=0.01kg{m_2} = 10g = 0.01kg
Now by using the formula we have already seen,
V=(m2em1m1+m2)u2+(m1em1m1+m2)u1V = \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
e=1\Rightarrow e = 1
On putting the values, we get
V=(0.011010+0.01)(1)+(101010+0.01)5\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
V=9.99+15010.01\Rightarrow V = \dfrac{{9.99 + 150}}{{10.01}}
Again solving the equation,
V=109.9910.01\Rightarrow V = \dfrac{{109.99}}{{10.01}}
And it will be approximately equal to
11m/s\Rightarrow \approx 11m/s
So if it is the case of elastic collision BB will move with the speed of approximately 11m/s11m/s.

Hence the option (c)\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.