A stationary particle breaks into two parts of masses \(m_A\) and \(m_B\), which move with velocities \(v_A\) and \(v_B\), respectively. The ratio of their kinetic energies (\(K_B : K_A\)) is:
- \(v_B : v_A\)
- \(m_B : m_A\)
- \(m_B v_B : m_A v_A\)
- \(1 : 1\)
The Correct Option is A
Approach Solution - 1
To solve this problem, we need to understand the concept of kinetic energy and the conservation of momentum in a system where a stationary particle breaks into two parts.
- The kinetic energy (\(K\)) of an object with mass \(m\) moving at velocity \(v\) is given by the formula: \(K = \frac{1}{2} m v^2\).
- Let's denote the kinetic energies of the two parts as \(K_A\) and \(K_B\), for masses \(m_A\) and \(m_B\) moving at velocities \(v_A\) and \(v_B\) respectively: \(K_A = \frac{1}{2} m_A v_A^2\) and \(K_B = \frac{1}{2} m_B v_B^2\).
- We need to find the ratio of the kinetic energies, \(K_B : K_A\). Thus:
\(\frac{K_B}{K_A} = \frac{\frac{1}{2} m_B v_B^2}{\frac{1}{2} m_A v_A^2} = \frac{m_B v_B^2}{m_A v_A^2}\)
- Applying the principle of conservation of momentum, where the total initial momentum is zero (since the particle is initially stationary), we have: \(m_A v_A = m_B v_B\).
- Using the above momentum relation, we can express: \(v_A = \frac{m_B v_B}{m_A}\).
- Substituting this into the kinetic energy ratio equation gives:
\(\frac{K_B}{K_A} = \frac{m_B v_B^2}{m_A \left(\frac{m_B v_B}{m_A}\right)^2} = \frac{v_B}{v_A}\)
- Therefore, the ratio of their kinetic energies is \(v_B : v_A\).
Approach Solution -2
Initial momentum is zero:
\[ P_A = P_B \implies m_A v_A = m_B v_B. \]
The kinetic energy ratio is:
\[ \frac{K_B}{K_A} = \frac{\frac{1}{2} m_B v_B^2}{\frac{1}{2} m_A v_A^2}. \]
\[ \frac{K_B}{K_A} = \frac{m_B v_B^2}{m_A v_A^2} = \frac{v_B}{v_A}. \]
Final Answer: vB : vA.
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 Kinetic Energy Questions
- The total kinetic energy of 1 mole of oxygen at 27°C is :
[Use universal gas constant (R)= 8.31 J/mole K] - Two bodies of mass 4 g and 25 g are moving with equal kinetic energies. The ratio of the magnitude of their linear momentum is:
- A uniform disc of mass \(5\) \(kg\) and radius \(2\) \(m\) is rotating with \(10\) \(rad/s\). Now another identical disc is gently placed on first disc. Because of friction both disc acquire common angular velocity. Loss of kinetic energy in process is ________J.
- An artillery of mass \(M_1\), fires a shell of mass \(M_2\). At the time of firing the ratio of kinetic energy is.
- When kinetic energy of a body becomes 36 times of its original value, the percentage increase in the momentum of the body will be
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 ______.