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Expert-verifiedA \(1.80 kg\) falcon catches a \(0.650 kg\) dove from behind in midair. What is their velocity after impact if the falcon’s velocity is initially \( 28.0 m/s\) and the dove’s velocity is \(7.00 m/s\) in the same direction?
The final velocity, \({v_f} = 22.4\,m/s\), is the velocity of the falcon and dove together, after the collision along the positive x-direction (that is in the direction motion of falcon and dove before the collision).
Given:
The initial velocity of the falcon, \({v_1} = 28.0 m/s\)
Mass of the falcon, \({m_1} = 1.80 kg\)
Mass of the dove, \({m_2} = 0.650 kg\)
The initial velocity of the dove, \({v_2} = 7.00 m/s\) ( v2 is positive because the dove is having the same direction as the falcon after catching the dove)
Total mass after the collision, \({m_t} = {m_1} + {m_2}\)
{Since the masses (falcon and dove) stick together after the collision}
The final velocity of the falcon = The final velocity of the dove = vf
The Law of conservation of momentum states that the total momentum of a system always remains constant before and after collisions or we can say that the initial momentum before the collision of a system is equal to the final momentum of the system after the collision.
Let p1 and p2 be the initial momentum of two objects before the collision, \({p'_1}\)and \({p'_2}\) be the final momentum after the collision, then according to the law of conservation of momentum,
\({p_1} + {p_2} = {p'_1} + {p'_2}\).................(1)
In this problem, after the collision, the two masses (mass of falcon and dove) move with a final velocity vf or the masses get to stick together. These types of collisions are known as inelastic collisions.
Therefore, equation 1 becomes\({p_1} + {p_2} = {p'_t}\).........(2)
Where \({p'_t}\) is the final momentum of the falcon and deer together.
Substituting the value of \(p = mv\) in equation 2, we get
\({m_1}{v_1} + {m_2}{v_2} = {m_t}{v_f}\).........(3) where vf is the final velocity of the dove and falcon together.
Dividing both sides of the equation 3 by mt, gives
\(\dfrac{{{m_1}{v_1} + {m_2}{v_2}}}{{{m_t}}} = {v_f}\)
Where vf is the final velocity of the system together.
\({v_f} = \dfrac{{{m_1}{v_1} + {m_2}{v_2}}}{{{m_t}}}............(4)
Substitute the values in equation 4, we get
\({v_f} = \dfrac{{(1.80kg \times 28m/s) + (0.650kg \times 7.00m/s)}}{{1.80kg + 0.650kg}}\)
\(\begin{aligned} &= \dfrac{{(1.80kg \times 28m/s) + (0.650kg \times 7.00m/s)}}{{1.80kg + 0.650kg}}\\ &= \dfrac{{(50.4kg \cdot m/s) + (4.55kg \cdot m/s)}}{{2.45kg}}\\ &= \dfrac{{54.95kg \cdot m/s}}{{2.45kg}}\\ &= 22.4m/s\end{aligned}\)
Thus the final velocity, \({v_f} = 22.4\,m/s\), is the velocity of the falcon and dove together, after the collision.
Hence, the final velocity is positive indicates that the system will move together along the positive x-direction after the collision or after catching the dove by the falcon.
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