Two bodies with masses \( m_1 \) and \( m_2 \) (\( m_1 > m_2 \)) are joined by a string passing over a fixed pulley. Assuming masses of the pulley and thread are negligible. Then the acceleration of the centre of mass of the system is:
1. \( \left(\frac{m_1 - m_2}{m_1 + m_2}\right) g \)
2. \( \left(\frac{m_1 - m_2}{m_1 + m_2}\right)^2 g \)
3. \( \frac{m_1 g}{(m_1 + m_2)} \)
4. \( \frac{m_2 g}{(m_1 + m_2)} \)
View Answer
The acceleration of each block is \( a = \left(\frac{m_1 - m_2}{m_1 + m_2}\right) g \). The acceleration of the center of mass is \( a_{\text{cm}} = \frac{m_1 a_1 + m_2 a_2}{m_1 + m_2} = \left(\frac{m_1 - m_2}{m_1 + m_2}\right) a = \left(\frac{m_1 - m_2}{m_1 + m_2}\right)^2 g \).
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:
1. \( 5\text{ m/s} \) away from the wall
2. \( 9\text{ m/s} \) away from the wall
3. \( 3\text{ m/s} \) away from the wall
4. \( 6\text{ m/s} \) away from the wall
View Answer
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
1. \( K = \left(\frac{m}{M}\right)^2 k \)
2. \( K = \sqrt{\frac{m}{M}} k \)
3. \( K = \left(\frac{m}{M}\right) k \)
4. \( K = \left(\frac{M}{m}\right) k \)
View Answer
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, \( K = \left(\frac{m}{M}\right) k \).
Given below are two statements:
Statement I: Centre of mass of any object always coincide with centre of gravity.
Statement II: Centre of gravity is the point where total gravitational torque on the body is zero.
In the light of the above statements, choose the most appropriate answer from the options given below.
1. Both statements I and II are correct
2. Both statements I and II are incorrect
3. Statement I is correct but II is incorrect
4. Statement I is incorrect but II is correct
View Answer
Statement I is false because the center of mass and center of gravity only coincide in a uniform gravitational field. Statement II is true because the center of gravity is defined as the point about which the net gravitational torque is zero.
A body of mass 2 kg moving with velocity \(5 \text{ m s}^{-1}\) collides with a body at rest of mass 3 kg and sticks to it. Now the combined mass starts moving. The final velocity of whole mass is
1. \(1 \text{ m s}^{-1}\)
2. \(2 \text{ m s}^{-1}\)
3. \(4 \text{ m s}^{-1}\)
4. \(3 \text{ m s}^{-1}\)
View Answer
Using the law of conservation of linear momentum: \(m_1 u_1 + m_2 u_2 = (m_1 + m_2) v_f\). Substituting the values: \(2(5) + 3(0) = (2 + 3) v_f ⇒ 10 = 5 v_f ⇒ v_f = 2 \text{ m s}^{-1}\).
Read the statements marked as assertion (A) and reason (R) and choose the correct option.
Assertion (A): If no external force acts on a system, the velocity of the centre of mass remains constant.
Reason (R): If there is no external force on system, then momentum of system is conserved.
1. Both (A) and (R) are true and (R) is the correct explanation of (A)
2. Both (A) and (R) are true but (R) is not the correct explanation of (A)
3. (A) is true but (R) is false
4. (A) is false but (R) is true
View Answer
If the net external force on a system is zero, its linear momentum is conserved (\(vec{P} = M\vec{v}_{\text{cm}} = \text{constant}\)). Consequently, the velocity of the centre of mass \(\vec{v}_{\text{cm}}\) remains constant.
Assertion (A): A body with negative energy cannot have linear momentum.
Reason (R): Magnitude of linear momentum can be negative.
1. Both (A) & (R) are true and the (R) is the correct explanation of the (A)
2. Both (A) & (R) are true but the (R) is not the correct explanation of the (A)
3. (A) is true but (R) is false
4. Both (A) and (R) are false
View Answer
Total mechanical energy \(E = K + U\) can be negative if potential energy \(U\) is negative and larger in magnitude than kinetic energy \(K\).
However, kinetic energy \(K = \frac{1}{2}mv^2\) is always non-negative, implying momentum exists. The magnitude of linear momentum \(|\vec{p}| = mv\) is always non-negative.
Therefore, both Assertion and Reason are false.