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.
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).
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 \(v_s = 100\text{ m/s}\).