Three bodies A, B and C have equal kinetic energies and their masses are 400 g, 1.2 kg and 1.6 kg respectively. The ratio of their linear momenta is :
- \(1 : \sqrt{3} : 2\)
- \(1 : \sqrt{3} : \sqrt{2}\)
- \(\sqrt{2} : \sqrt{3} : 1\)
- \(\sqrt{3} : \sqrt{2} : 1\)
The Correct Option is A
Approach Solution - 1
To solve the problem of finding the ratio of linear momenta for three bodies with equal kinetic energies, we start by considering the relationship between kinetic energy and momentum.
The kinetic energy (\(KE\)) of a body is given by the formula:
\(KE = \frac{1}{2}mv^2\)
where \(m\) is the mass and \(v\) is the velocity of the body.
The linear momentum (\(p\)) of a body is given by:
\(p = mv\)
We are told that the kinetic energies of bodies A, B, and C are equal. Therefore, for each body:
\(\frac{1}{2} m_A v_A^2 = \frac{1}{2} m_B v_B^2 = \frac{1}{2} m_C v_C^2\)
Since the kinetic energies are equal, we can write:
\(m_A v_A^2 = m_B v_B^2 = m_C v_C^2\)
Now, we can express the momentum for each body in terms of its velocity:
\(v_A = \sqrt{\frac{2 \times KE}{m_A}}\), \(v_B = \sqrt{\frac{2 \times KE}{m_B}}\), \(v_C = \sqrt{\frac{2 \times KE}{m_C}}\)
Then, the momentum for each body is:
\(p_A = m_A \times v_A = m_A \times \sqrt{\frac{2 \times KE}{m_A}}\), \(p_B = m_B \times v_B = m_B \times \sqrt{\frac{2 \times KE}{m_B}}\), \(p_C = m_C \times v_C = m_C \times \sqrt{\frac{2 \times KE}{m_C}}\)
Simplifying further, we find the momentum expressions:
\(p_A = \sqrt{2 \times KE \times m_A}\), \(p_B = \sqrt{2 \times KE \times m_B}\), \(p_C = \sqrt{2 \times KE \times m_C}\)
Given masses are \(m_A = 400\) g, \(m_B = 1.2\) kg, and \(m_C = 1.6\) kg. Converting \(m_A\) to kg, we have \(m_A = 0.4\) kg.
Using these masses to find the ratio of their momentum:
Momentum ratio:
\(\frac{p_A}{p_B} = \frac{\sqrt{0.4}}{\sqrt{1.2}} = \frac{\sqrt{1}}{\sqrt{3}} = \frac{1}{\sqrt{3}}\)
\(\frac{p_B}{p_C} = \frac{\sqrt{1.2}}{\sqrt{1.6}} = \frac{\sqrt{3}}{\sqrt{4}} = \frac{\sqrt{3}}{2}\)
Solving for the combined ratio gives:
\(p_A : p_B : p_C = 1 : \sqrt{3} : 2\)
Thus, the correct option is \(1 : \sqrt{3} : 2\).
Approach Solution -2
The kinetic energy is given by:
\[ KE = \frac{P^2}{2m}, \quad \text{where } P \propto \sqrt{m}. \]
For masses \(m_A = 400 \, \text{g}, \, m_B = 1.2 \, \text{kg}, \, m_C = 1.6 \, \text{kg}\):
\[ P_A : P_B : P_C = \sqrt{400} : \sqrt{1200} : \sqrt{1600}. \]
Simplify:
\[ P_A : P_B : P_C = 1 : \sqrt{3} : 2. \]
Final Answer: 1 : \(\sqrt{3}\) : 2.
Top JEE Main Physics Questions
- In a hydrogen like atom, when an electron jumps from the $M$ - shell to the $L$ - shell, the wavelength of emitted radiation is $\lambda$. If an electron jumps from $N$-shell to the $L$-shell, the wavelength of emitted radiation will be :
- Two masses $m$ and $\frac{m}{2}$ are connected at the two ends of a massless rigid rod of length $l$. The rod is suspended by a thin wire of torsional constant $k$ at the centre of mass of the rod-mass system(see figure). Because of torsional constant $k$, the restoring torque is $\tau =k\theta$ for angular displacement $\theta$. If the rod is rota ted by $\theta_0$ and released, the tension in it when it passes through its mean position will be:
A black body is at a temperature of 2880 K. The energy of radiation emitted by this body with wavelength between 499 nm and 500 nm is U1, between 999 nm and 1000 nm is U2 and between 1499 nm and 1500 nm is U3. The Wien's constant, b = 2.88×106 nm-K. Then,
- $I _{ CM }$ is the moment of inertia of a circular disc about an axis (CM) passing through its center and perpendicular to the plane of disc $I_{A B}$ is it's moment of inertia about an axis $AB$ perpendicular to plane and parallel to axis $CM$ at a distance $\frac{2}{3} R$ from center Where $R$ is the radius of the dise The ratio of $I _{ AB }$ and $I _{ CM }$ is $x: 9$ The value of $x$ is______
- A nucleus disintegrates into two smaller parts, which have their velocities in the ratio $3: 2$ The ratio of their nuclear sizes will be $\left(\frac{x}{3}\right)^{\frac{1}{3}}$ The value of ' $x$ ' is:-
Top JEE Main Momentum and Kinetic Energy in Collisions Questions
- A body of mass 2 kg moving with a speed of 4 m/s makes an elastic collision with another body at rest and continues to move in the original direction but with one fourth of its initial speed. The speed of the two body centre of mass is $\frac{x}{10}$ m/s. Then the value of $x$ is ________.
- Two masses of 3.4 kg and 2.5 kg are accelerated from an initial speed of 5 m/s and 12 m/s, respectively. The distances traversed by the masses in the 5th second are 104 m and 129 m, respectively. The ratio of their momenta after 10 s is \(\frac{x}{8}\). The value of \(x\) is ________.
Top JEE Main Questions
- In a hydrogen like atom, when an electron jumps from the $M$ - shell to the $L$ - shell, the wavelength of emitted radiation is $\lambda$. If an electron jumps from $N$-shell to the $L$-shell, the wavelength of emitted radiation will be :
- Two masses $m$ and $\frac{m}{2}$ are connected at the two ends of a massless rigid rod of length $l$. The rod is suspended by a thin wire of torsional constant $k$ at the centre of mass of the rod-mass system(see figure). Because of torsional constant $k$, the restoring torque is $\tau =k\theta$ for angular displacement $\theta$. If the rod is rota ted by $\theta_0$ and released, the tension in it when it passes through its mean position will be:
What will be the equilibrium constant of the given reaction carried out in a \(5 \,L\) vessel and having equilibrium amounts of \(A_2\) and \(A\) as \(0.5\) mole and \(2 \times 10^{-6}\) mole respectively?
The reaction : \(A_2 \rightleftharpoons 2A\)- The value of is
- The number of terms of an is even. The sum of the odd terms is and of the even terms is . The last term exceeds the first by . Then the number of terms in the series is ______.