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 \(m_3 = p_3/v_3 = 20/4 = 5\text{ kg}\).
A shell of mass \(200\text{ gm}\) is ejected from a gun of mass \(4\text{ kg}\) by an explosion that generates \(1.05\text{ kJ}\) of energy. The initial velocity of the shell is
(2008)
1. \(40\text{ m/s}\)
2. \(120\text{ m/s}\)
3. \(100\text{ m/s}\)
4. \(80\text{ m/s}\)
View Answer
Let shell mass \(m_s = 0.2\text{ kg}\), gun mass \(m_g = 4\text{ kg}\). Energy \(E = 1050\text{ J}\). By momentum conservation \(m_s v_s = m_g v_g\), so \(v_g = \frac{m_s v_s}{m_g} = \frac{0.2 v_s}{4} = \frac{v_s}{20}\). The energy is KE: \(E = \frac{1}{2}m_s v_s^2 + \frac{1}{2}m_g v_g^2 = \frac{1}{2}(0.2)v_s^2 + \frac{1}{2}(4)(\frac{v_s}{20})^2 = 0.1v_s^2 + \frac{2v_s^2}{400} = 0.1v_s^2 + 0.005v_s^2 = 0.105v_s^2\). Thus, \(v_s^2 = \frac{1050}{0.105} = 10000\), so \(v_s = 100\text{ m/s}\).
A metal ball of mass $2\text{ kg}$ moving with speed of $36\text{ km/h}$ has a head on collision with a stationary ball of mass $3\text{ kg}$. If after collision, both the balls move as a single mass, then the loss in K.E. due to collision is:
(1997)
1. $100\text{ J}$
2. $140\text{ J}$
3. $40\text{ J}$
4. $60\text{ J}$
View Answer
Initial kinetic energy $K_i = \frac{1}{2}m_1 u_1^2 = 100\text{ J}$ (with $u_1 = 10\text{ m/s}$). Final velocity $v = \frac{m_1 u_1}{m_1+m_2} = 4\text{ m/s}$, and final kinetic energy $K_f = \frac{1}{2}(m_1+m_2)v^2 = 40\text{ J}$. Loss in K.E. = $100 - 40 = 60\text{ J}$.