Center of Mass , Momentum and Collision - NEET Physics Questions
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Center of Mass , Momentum and Collision

Practice Questions:
Question 51: moderate

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 \(m_3 = p_3/v_3 = 20/4 = 5\text{ kg}\).

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Question 52: moderate

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

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