Masses 2 kg and 5kg are located at origin and (7,0) respectively. Centre of mass of the arrangement is located at
Centre of mass is the point which divides the line joining the masses in reverse ratio of masses.
Here ratio of mass is 2:5 so distance from 2kg mass located at origin will be 5/7 of 7 cm so
\[ x_{cm}= \frac{2(0)+5(7)}{5+2}= 5 \]
In carbon monoxide molecules, the carbon and the oxygen atoms are separated by a distance of 1.2 × 10 –10 m. The distance of the centre of mass from the carbon atom is
\[ x_{cm}= \frac{(m_{1}x_{1} + m_{2}x_{2})}{m_{1}+m_{2}} \]
\[ x_{cm}= \frac{12(0)+ 16 (1.2 × 10^{–10})}{12+16}= 0.69 \times 10^{–10} m \]
Two particles \(A\) and \(B\) initially at rest, move towards each other under mutual force of attraction. At an instance when the speed of \(A\) is \(v\) and speed of \(B\) is \(3v\), the speed of centre of mass is
Since there is no external force acting on the system, the acceleration of the centre of mass is zero. Since the system was initially at rest, the velocity of the centre of mass remains zero.
A bullet of mass \(m\) hits a block of mass \(M\) elastically. The transfer of energy is the maximum, when
In a one-dimensional elastic collision between a moving body of mass \(m\) and a stationary body of mass \(M\), maximum transfer of kinetic energy occurs when the masses are equal (\(m = M\)).
A stationary bomb explodes into two parts, \(4\text{ kg}\) and \(8\text{ kg}\). The velocity of the \(8\text{ kg}\) mass is \(6\text{ ms}^{-1}\). The KE of the other body is:
By conservation of momentum, \(m_1 v_1 = m_2 v_2\), which gives \(4 \times v_1 = 8 \times 6\), so \(v_1 = 12\text{ ms}^{-1}\). The kinetic energy of the \(4\text{ kg}\) body is \(K = \frac{1}{2} m_1 v_1^2 = \frac{1}{2} \times 4 \times 12^2 = 288\text{ J}\).
A stationary body of mass m explodes into 3 parts with mass ratio of \(1 : 3 : 3\). The two fragments with equal mass move at right angles to each other with velocity of \(15\text{ ms}^{-1}\). The velocity of the third fragment is (in \(\text{ms}^{-1}\)):
The ratio of masses is \(m' : 3m' : 3m'\). The combined momentum of the two perpendicular \(3m'\) masses is \(P = \sqrt{(3m' \times 15)^2 + (3m' \times 15)^2} = 45\sqrt{2} m'\). Conservation of momentum requires the third fragment \(m'\) to balance this: \(m' v_3 = 45\sqrt{2} m' ⇒ v_3 = 45\sqrt{2}\text{ ms}^{-1}\).
A body with kinetic energy K moving in +X direction splits up into two parts A and B of equal mass on its own. Part ‘A’ moves back in -X direction with a velocity equal in magnitude to the initial velocity of the body. The kinetic energy of part B will be:
Let the total mass be \(2m\), so \(K = mv_0^2\). Under momentum conservation, \(2mv_0 = m(-v_0) + mv_B ⇒ v_B = 3v_0\). The kinetic energy of part B is \(K_B = \frac{1}{2}m(3v_0)^2 = \frac{9}{2}mv_0^2 = \frac{9}{2}K\).
A stationary object explodes into two parts of equal masses, then:
Assertion: Both parts will have same kinetic energy after explosion.
Reason : Both parts will have same momentum after explosion.
By conservation of momentum, the two parts move in opposite directions with equal magnitude of momentum, \( \vec{p}_1 = -\vec{p}_2 \). Thus, they have different momenta (since momentum is a vector), so the Reason is false. However, since they have the same mass and same momentum magnitude, their kinetic energy \( K = \frac{p^2}{2m} \) is the same, so the Assertion is true.
Assertion: If kinetic energy of a system of particles is zero, then linear momentum of system must be zero.
Reason: If linear momentum of a system of particles is zero, then kinetic energy of system must be zero.
If the kinetic energy of a system is zero, the speed of each particle must be zero, meaning the total linear momentum is also zero. If the total linear momentum is zero, particles can still be moving in opposite directions, resulting in a non-zero kinetic energy. Thus, the Assertion is true but the Reason is false.
A particle is dropped on the ground from a height \( 36\text{ m} \) and after striking the ground it rebounds to a height of \( 16\text{ m} \). Then the co-efficient of restitution of the collision with the ground is:
The coefficient of restitution \( e \) for a vertical rebound is given by \( e = \sqrt{\frac{h_{\text{rebound}}}{h_{\text{initial}}}} \). Substituting the given values, \( e = \sqrt{\frac{16}{36}} = \frac{4}{6} = \frac{2}{3} \).