A ball of mass 4 kg moving on a smooth horizontal surface makes an elastic collision with another ball of mass m at rest in the line of motion of the first ball. If after collision first ball moves in the same direction with one fourth of its velocity before collision, then the mass of second ball is:
(1) 4 kg
(2) 4.4 kg
(3) 2.4 kg
(4) 2 kg
Answer
594.3k+ views
Hint: Since it is an elastic collision, both the energy and momentum will be conserved. So, from the conservation of linear momentum equation, we can reach the velocity of the first ball and from that, we can find the mass of the second ball. This will help you in answering this question.
Complete answer:
According to the law of conservation of linear momentum, we can write that,
\[{{m}_{1}}{{u}_{1}}+{{m}_{2}}{{u}_{2}}={{m}_{1}}{{v}_{1}}+{{m}_{2}}{{v}_{2}}\]
Where \[{{m}_{1}}\]be the mass of the first ball, \[{{m}_{2}}\] be the mass of the second ball, \[{{u}_{1}}\]be the initial velocity of first ball, \[{{u}_{2}}\] be the initial velocity of second ball, \[{{v}_{1}}\] be the final velocity of first ball and \[{{v}_{2}}\] be the final velocity of the second ball.
From the above equation rearranging its form, we will get the initial velocity as follows,
\[{{v}_{1}}=\dfrac{({{m}_{1}}-{{m}_{2}})}{({{m}_{1}}+{{m}_{2}})}{{u}_{1}}+\left( \dfrac{2{{m}_{2}}}{{{m}_{1}}+{{m}_{2}}} \right){{u}_{2}}\]
Since the second ball is at rest, the equation will become,
\[{{v}_{1}}=\dfrac{({{m}_{1}}-{{m}_{2}})}{({{m}_{1}}+{{m}_{2}})}{{u}_{1}}\]
It is given that,
\[{{v}_{2}}=\dfrac{{{v}_{1}}}{4}\]
So, substituting values in the final equation, we will be getting like this,
\[\begin{align}
& \dfrac{1}{4}=\dfrac{4-m}{4+m} \\
& \Rightarrow 4+m=16-4m \\
& \Rightarrow 5m=12 \\
& \Rightarrow m=\dfrac{12}{5}\text{ }kg \\
& \therefore m=2.4\text{ }kg \\
\end{align}\]
Hence option (3) will be the answer. The mass of the second ball will be \[2.4\text{ }kg\].
Note:
The major objective of study of collision is to find out as much as possible about the forces that act during a collision from the knowledge of the state of particles before and after the collision.in fact all our understanding of the subatomic world-electrons, protons, quarks, comes from the experiments involving collisions.
Complete answer:
According to the law of conservation of linear momentum, we can write that,
\[{{m}_{1}}{{u}_{1}}+{{m}_{2}}{{u}_{2}}={{m}_{1}}{{v}_{1}}+{{m}_{2}}{{v}_{2}}\]
Where \[{{m}_{1}}\]be the mass of the first ball, \[{{m}_{2}}\] be the mass of the second ball, \[{{u}_{1}}\]be the initial velocity of first ball, \[{{u}_{2}}\] be the initial velocity of second ball, \[{{v}_{1}}\] be the final velocity of first ball and \[{{v}_{2}}\] be the final velocity of the second ball.
From the above equation rearranging its form, we will get the initial velocity as follows,
\[{{v}_{1}}=\dfrac{({{m}_{1}}-{{m}_{2}})}{({{m}_{1}}+{{m}_{2}})}{{u}_{1}}+\left( \dfrac{2{{m}_{2}}}{{{m}_{1}}+{{m}_{2}}} \right){{u}_{2}}\]
Since the second ball is at rest, the equation will become,
\[{{v}_{1}}=\dfrac{({{m}_{1}}-{{m}_{2}})}{({{m}_{1}}+{{m}_{2}})}{{u}_{1}}\]
It is given that,
\[{{v}_{2}}=\dfrac{{{v}_{1}}}{4}\]
So, substituting values in the final equation, we will be getting like this,
\[\begin{align}
& \dfrac{1}{4}=\dfrac{4-m}{4+m} \\
& \Rightarrow 4+m=16-4m \\
& \Rightarrow 5m=12 \\
& \Rightarrow m=\dfrac{12}{5}\text{ }kg \\
& \therefore m=2.4\text{ }kg \\
\end{align}\]
Hence option (3) will be the answer. The mass of the second ball will be \[2.4\text{ }kg\].
Note:
The major objective of study of collision is to find out as much as possible about the forces that act during a collision from the knowledge of the state of particles before and after the collision.in fact all our understanding of the subatomic world-electrons, protons, quarks, comes from the experiments involving collisions.
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