Linear Momentum and Second Law of Motion: Practice Problem & Solution
A particle is projected with velocity \( u \) makes an angle \( \theta \) w.r.t. horizontal. Now it breaks in two identical parts at highest point of trajectory. If one part is retrace its path, then velocity of other part is: (1999)
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 \( u \cos\theta \) (horizontal) and mass is \( M \). Initial momentum \( P_i = M u \cos\theta \). Particle breaks into two identical parts (\( M/2 \) each). One part retraces its path, so its velocity is \( -u \cos\theta \). By conservation of momentum: \( M u \cos\theta = (M/2) (-u \cos\theta) + (M/2) v_2 \). Solving for \( v_2 \) gives \( v_2 = 3u \cos\theta \).
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