Linear Momentum and Second Law of Motion: Practice Problem & Solution
A projectile is thrown at a speed of \(100\text{ m/s}\) at an angle of \(37^\circ\) above the horizontal. At the highest point the projectile breaks into two parts of mass ratio \(1 : 3\), the smaller coming to rest. Find the speed of the second piece.
Solution Explained:
To solve this problem, we apply the core principles of Linear Momentum and Second Law of Motion. Understanding the underlying formula is key to arriving at the correct answer below:
At the highest point, velocity is horizontal: \(v_x = u\cos(37^\circ) = 100 \times 0.8 = 80\text{ m/s}\). By conservation of linear momentum along the horizontal: \(m v_x = m_1 v_1 + m_2 v_2\). Here \(m_1 = m/4\) (comes to rest, \(v_1 = 0\)) and \(m_2 = 3m/4\). Thus, \(m(80) = \frac{3m}{4} v_2\), which gives \(v_2 = \frac{320}{3}\text{ m/s}\).
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