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.
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, \( e = \sqrt{\frac{16}{36}} = \frac{4}{6} = \frac{2}{3} \).
A ball of mass \( m \) approaches a wall of mass \( M \) (\( M \gg m \)) with speed \( 4\text{ m/s} \) along the normal to the wall. The speed of the wall is \( 1\text{ m/s} \) towards the ball. The speed of the ball after an elastic collision with the wall is:
Using the coefficient of restitution \( e = 1 \), the relative velocity of separation equals the relative velocity of approach. The approach velocity is \( 4 - (-1) = 5\text{ m/s} \). After collision, the relative separation velocity is \( v' - 1 = 5 \implies v' = 6\text{ m/s} \) away from the wall.
A bullet of mass \( m \) leaves the barrel of a gun of mass \( M \) with a velocity \( v \). The gun is known to recoil with a velocity \( V \). If \( k \) and \( K \) respectively denote the kinetic energies of the bullet and the gun respectively; then
By conservation of momentum, the bullet and gun have equal momentum magnitude, \( p \). Since kinetic energy is \( K_{\text{E}} = \frac{p^2}{2\text{mass}} \), we have \( k = \frac{p^2}{2m} \) and \( K = \frac{p^2}{2M} \). Thus, \( K = \left(\frac{m}{M}\right) k \).
A block of mass \( m \) moving with a velocity \( v \) collides with another block of mass \( M \) at rest. The two blocks stick together due to the collision. The loss of K.E. expressed as a fraction of total initial kinetic energy is:
By conservation of momentum, the final velocity after a completely inelastic collision is \( v_f = \frac{mv}{m+M} \). The fractional loss of kinetic energy is \( \frac{K_i - K_f}{K_i} = 1 - \frac{\frac{1}{2}(m+M)v_f^2}{\frac{1}{2}mv^2} = \frac{M}{m+M} \).
Assertion (A): In any kind of collision, kinetic energy cannot be same throughout.
Reason (R): In elastic collision kinetic energy remains constant throughout.
Assertion (A) is true as kinetic energy is not conserved in all types of collisions (e.g., inelastic collisions). Reason (R) is true as kinetic energy is conserved in elastic collisions. However, (R) does not correctly explain (A).