Assertion (A): In a perfectly inelastic collision there is a limit to the loss of kinetic energy of colliding bodies.
Reason (R): In perfectly inelastic collision, linear momentum of system is conserved.
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
Both (A) and (R) are true. In a perfectly inelastic collision, momentum is conserved (R), which allows the calculation of the final common velocity (\(v_f = \frac{m_1 v_1 + m_2 v_2}{m_1 + m_2}\)) and thus the minimum kinetic energy (\(KE_f = \frac{1}{2}(m_1+m_2)v_f^2\)) that must remain, placing a limit on kinetic energy loss (A). Hence, (R) correctly explains (A).
Consider a one-dimensional head on collision of two balls.
Assertion (A): The loss in kinetic energy of the system during the collision does not depend on the velocity of the observer.
Reason (R): Kinetic energy of a body is independent of velocity of observer.
1. (1) Both (A) & (R) are true and the (R) is the correct explanation of the (A)
2. (2) Both (A) & (R) are true but the (R) is not the correct explanation of the (A)
3. (3) (A) is true but (R) is false
4. (4) Both (A) and (R) are false
View Answer
Assertion (A): The loss in kinetic energy of a system is generally dependent on the observer's frame of reference. Thus, (A) is false.
Reason (R): Kinetic energy \(K = \frac{1}{2}mv^2\) depends on the velocity (v\), which is relative to the observer. Therefore, kinetic energy is dependent on the velocity of the observer. Thus, (R) is false.
Since both (A) and (R) are false, option (4) is correct.
Assertion (A): When one object collides with another object, the impulse during deformation and reformation will be in same direction on one particular object.
Reason (R): Due to deformation impulse the objects first deform and due to the same reformation impulse, they again try to regain its original shape.
1. (1) Both (A) & (R) are true and the (R) is the correct explanation of the (A)
2. (2) Both (A) & (R) are true but the (R) is not the correct explanation of the (A)
3. (3) (A) is true but (R) is false
4. (4) Both (A) and (R) are false
View Answer
Assertion (A): During collision, the impulses during the deformation phase and reformation phase on a particular object act in opposite directions. So, (A) is false.
Reason (R): The deformation impulse and reformation impulse are distinct. They are not the 'same' impulse. So, (R) is false. Since both (A) and (R) are false, option (4) is correct.
Assertion (A): Maximum energy loss occurs when the particles get stuck together as a result of collision.
Reason (R): A point particle of mass (m\) moving with speed (v\) collides with stationary point particle of mass (M\). Then the maximum energy loss possible is given \( \frac{m}{(m+M)}\left(\frac{1}{2}mv^2\right)\).
1. (1) Both (A) & (R) are true and the (R) is the correct explanation of the (A)
2. (2) Both (A) & (R) are true but the (R) is not the correct explanation of the (A)
3. (3) (A) is true but (R) is false
4. (4) Both (A) and (R) are false
View Answer
Assertion (A): Maximum kinetic energy loss occurs in a perfectly inelastic collision where particles stick together. So, (A) is true.
Reason (R): For a perfectly inelastic collision between mass (m\) (velocity (v\)) and stationary mass (M\), the energy loss is ( \Delta K = \frac{M}{(m+M)}\left(\frac{1}{2}mv^2\right)\). The given formula in (R) is incorrect.
So, (R) is false. Therefore, (A) is true and (R) is false. Option (3) is correct.
Assertion (A): In case of bullet fired from a gun, the ratio of kinetic energy of gun and bullet is equal to ratio of masses of bullet and gun.
Reason (R): In firing of bullet, linear momentum of system is conserved.
1. (1) Both (A) & (R) are true and the (R) is the correct explanation of the (A)
2. (2) Both (A) & (R) are true but the (R) is not the correct explanation of the (A)
3. (3) (A) is true but (R) is false
4. (4) Both (A) and (R) are false
View Answer
Reason (R): For the bullet-gun system, the forces causing the bullet to fire are internal. Thus, linear momentum of the system is conserved. So, (R) is true.
Assertion (A): Let (m\) and (M\) be masses of bullet and gun, (v\) and (V\) their velocities. By momentum conservation, (mv = MV\). The ratio of kinetic energies is \( \frac{K_g}{K_b} = \frac{\frac{1}{2}MV^2}{\frac{1}{2}mv^2} = \frac{M(mv/M)^2}{mv^2} = \frac{m}{M}\). So, (A) is true.
(R) correctly explains (A) as the kinetic energy ratio is derived directly from momentum conservation. Option (1) is correct.
Consider the given statements and choose the correct option that follows:
Statement 1: During a collision the total linear momentum of system is conserved at each instant of collision.
Statement 2: During a collision the kinetic energy conservation holds always.
Based on above information, pick the correct option.
1. Both statements (1) and (2) are true
2. Both statements (1) and (2) are false
3. Statement (1) is true but (2) is false
4. Statement (1) is false but (2) is true
View Answer
Total linear momentum is conserved at each instant of collision because no external forces act. Kinetic energy, however, is not conserved during the period of deformation, and is conserved after only in perfectly elastic collisions. Thus, Statement 1 is true and Statement 2 is false.
An explosion blows a rock into three parts. Two parts go off at right angles to each other. These two are, \( 1 \text{ kg} \) first part moving with a velocity of \( 12 \text{ m s}^{-1} \) and \( 2 \text{ kg} \) second part moving with a velocity of \( 8 \text{ m s}^{-1} \). If the third part flies off with a velocity of \( 4 \text{ m s}^{-1} \), its mass would be:
(2009)
1. \( 7 \text{ kg} \)
2. \( 17 \text{ kg} \)
3. \( 3 \text{ kg} \)
4. \( 5 \text{ kg} \)
View Answer
By conservation of momentum, \( P_{text{total}} = 0 \). Momentum of first part \( P_1 = 1 \text{ kg} \times 12 \text{ m/s} = 12 \text{ Ns} \). Momentum of second part \( P_2 = 2 \text{ kg} \times 8 \text{ m/s} = 16 \text{ Ns} \). As \( P_1 \) and \( P_2 \) are perpendicular, their resultant \( P_{12} = sqrt{12^2 + 16^2} = 20 \text{ Ns} \). For conservation, \( P_3 \) must be \( 20 \text{ Ns} \). \( m_3 = P_3 / v_3 = 20 \text{ Ns} / 4 \text{ m/s} = 5 \text{ kg} \).
A mass of \( 1 \text{ kg} \) is thrown up with a velocity of \( 100 \text{ m/s} \). After \( 5 \) seconds, it explodes into two parts. One part of mass \( 400 \text{ g} \) comes down with a velocity \( 25 \text{ m/s} \). Calculate the velocity of other part:
(2000)
1. \( 40 \text{ m/s} \) upward
2. \( 40 \text{ m/s} \) downward
3. \( 100 \text{ m/s} \) upward
4. \( 60 \text{ m/s} \) downward
View Answer
Velocity of \( 1 \text{ kg} \) mass after \( 5 \text{ s} \): \( v = u - gt = 100 - 10 \times 5 = 50 \text{ m/s} \) (upward). Initial momentum before explosion \( P_i = 1 \text{ kg} \times 50 \text{ m/s} = 50 \text{ Ns} \) (upward). Mass of first part \( m_1 = 0.4 \text{ kg} \), \( v_1 = -25 \text{ m/s} \). Mass of second part \( m_2 = 0.6 \text{ kg} \). By conservation of momentum: \( P_i = m_1 v_1 + m_2 v_2 \). \( 50 = 0.4 \times (-25) + 0.6 v_2 \). \( 50 = -10 + 0.6 v_2 \implies v_2 = 100 \text{ m/s} \) (upward).
A body of mass \(4m\) is lying in \(x-y\) plane at rest. It suddenly explodes into three pieces. Two pieces each of mass \(m\) move perpendicular to each other with equal speeds \(v\). The total kinetic energy generated due to explosion is:
(2014)
1. \(mv^2\)
2. \(3/2 mv^2\)
3. \(2 mv^2\)
4. \(4 mv^2\)
View Answer
Initial momentum is zero. Two pieces of mass \(m\) move with velocity \(v\) perpendicular to each other. Their momenta are \(m\vec{v}_1 = mv\hat{i}\, m\vec{v}_2 = mv\hat{j}\). The third piece has mass \(m_3 = 4m - m - m = 2m\). By momentum conservation, \(m_3\vec{v}_3 = -(mv\hat{i} + mv\hat{j})\), so \(|\vec{v}_3| = \frac{\sqrt{(mv)^2 + (mv)^2}}{2m} = \frac{\sqrt{2}mv}{2m} = \frac{v}{\sqrt{2}}\). Total KE = \(\frac{1}{2}mv^2 + \frac{1}{2}mv^2 + \frac{1}{2}(2m)(\frac{v}{\sqrt{2}})^2 = mv^2 + \frac{1}{2}mv^2 = \frac{3}{2}mv^2\).
An explosion breaks a rock into three parts in a horizontal plane. Two of them go off at right angles to each other. The first part of mass \(1\text{ kg}\) moves with a speed of \(12\text{ ms}^{-1}\) and the second part of mass \(2\text{ kg}\) moves with \(8\text{ ms}^{-1}\) speed. If the third part flies off with \(4\text{ ms}^{-1}\) speed, then its mass is:
(2013, 2009)
1. \(17\text{ kg}\)
2. \(3\text{ kg}\)
3. \(5\text{ kg}\)
4. \(7\text{ kg}\)
View Answer
By conservation of momentum, the initial momentum is zero. Momentum of first part \(p_1 = 1\text{ kg} \times 12\text{ m/s} = 12\text{ kg m/s}\). Momentum of second part \(p_2 = 2\text{ kg} \times 8\text{ m/s} = 16\text{ kg m/s}\). Since they are perpendicular, resultant momentum \(p_{12} = \sqrt{12^2 + 16^2} = \sqrt{144 + 256} = \sqrt{400} = 20\text{ kg m/s}\). The third part must have momentum \(p_3 = 20\text{ kg m/s}\). Given its speed \(v_3 = 4\text{ m/s}\), its mass \(m_3 = p_3/v_3 = 20/4 = 5\text{ kg}\).