Question

If \( m \) is the mass of the projectile thrown upwards with velocity \( u \) at an angle \( \theta \) with the ground, then the change in momentum from the lowest to highest point of its trajectory will be

• A. \( mu \)
• B. \( 2mu \)
• C. \( mu \sin \theta \)
• D. \( mu \cos \theta \)

Hint:

The change in momentum during projectile motion depends on the vertical component of the velocity, as the horizontal component remains constant.

Answer 1

The Correct Option is C

Step 1: Understand momentum at different points.
Momentum \( p \) is given by: \[ p = mv, \] where \( m \) is the mass and \( v \) is the velocity.
Step 2: Analyze velocity at lowest and highest points.
At the highest point of the trajectory, the vertical component of the velocity is zero, so only the horizontal component remains. The horizontal component of the velocity is \( u \cos \theta \). At the lowest point, the velocity is \( u \sin \theta \).
Step 3: Calculate the change in momentum.
The change in momentum is the difference between the momentum at the highest point and the lowest point. Since the horizontal component of velocity is constant, the change in momentum only depends on the vertical component: \[ \Delta p = m \cdot (u \sin \theta). \]
Step 4: Conclusion.
Thus, the change in momentum is \( mu \sin \theta \), which corresponds to option (C).