
A \[750g\] peregrine falcon dives straight down towards a \[400g\] pigeon, which is flying level to the ground. Just before the falcon makes impact its velocity is $35m/s$. The velocity of the falcon and the pigeon in its talons immediately after impact is most nearly
\[\begin{align}
& A.35m/s \\
& B.31.95m/s \\
& C.28.9m/s \\
& D.25.85m/s \\
& E.22.8m/s \\
\end{align}\]
Answer
573.3k+ views
Hint: Here, there is a collision between the falcon and pigeon. Assuming the collision is elastic i.e. there is no loss in energy and that the velocity after collision of the birds is the same irrespective of the masses of the birds, we can find the velocity of the birds after collisions.
Formula used:
$mu=m’ v$
Complete step-by-step answer:
Given that the mass of the falcon is $m_{f}=750g$ is flying with a initial velocity $u_{f}=35m/s$ and the mass of the pigeon is $m_{p}=400g$. Let us assume that the pigeon is at rest initially, $u_{p}=0m/s$. Let us assume that velocity after collision is $v$, and is the same irrespective of the masses of the birds.
Assume that the collision is elastic, and then we know that the momentum of the system is conserved.
It is given mathematically as, $mu=m\prime v$ where $m$ and $u$ are the mass and the velocity of the system before collision and $m\prime$ and $v$ are the mass and the velocity of the system after collision .
Here, the falcon and the pigeon are considered as a system. We can then say that,
$m_{p}u_{p}+m_{f}u_{f}=m_{f}v+m_{p}v$
Substituting the values, we get, $m_{f}u_{f}=(m_{f}+m_{p})v$ as $u_{p}=0$
Substituting the values, we get, $750\times 35=(750+400)\times v$
Rearranging and solving we get, $v=\dfrac{26250}{1150}=22.8m/s$
Hence the answer is \[E.22.8m/s\]
So, the correct answer is “Option E”.
Note: Momentum denotes the state of motion or rest of any object. It is given as the product of the mass and the velocity of the object.$mu=m\prime v$, here the mass of the system after collision is denoted as $m\prime$, as after collision the mass of the system varies like the mass of the reactants add up or cancel, but the product of the mass and the velocity remains constant. In an elastic collision the total kinetic energy of the system is conserved.
Formula used:
$mu=m’ v$
Complete step-by-step answer:
Given that the mass of the falcon is $m_{f}=750g$ is flying with a initial velocity $u_{f}=35m/s$ and the mass of the pigeon is $m_{p}=400g$. Let us assume that the pigeon is at rest initially, $u_{p}=0m/s$. Let us assume that velocity after collision is $v$, and is the same irrespective of the masses of the birds.
Assume that the collision is elastic, and then we know that the momentum of the system is conserved.
It is given mathematically as, $mu=m\prime v$ where $m$ and $u$ are the mass and the velocity of the system before collision and $m\prime$ and $v$ are the mass and the velocity of the system after collision .
Here, the falcon and the pigeon are considered as a system. We can then say that,
$m_{p}u_{p}+m_{f}u_{f}=m_{f}v+m_{p}v$
Substituting the values, we get, $m_{f}u_{f}=(m_{f}+m_{p})v$ as $u_{p}=0$
Substituting the values, we get, $750\times 35=(750+400)\times v$
Rearranging and solving we get, $v=\dfrac{26250}{1150}=22.8m/s$
Hence the answer is \[E.22.8m/s\]
So, the correct answer is “Option E”.
Note: Momentum denotes the state of motion or rest of any object. It is given as the product of the mass and the velocity of the object.$mu=m\prime v$, here the mass of the system after collision is denoted as $m\prime$, as after collision the mass of the system varies like the mass of the reactants add up or cancel, but the product of the mass and the velocity remains constant. In an elastic collision the total kinetic energy of the system is conserved.
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