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).

Answer 2

Instead of comparing velocities at two points directly, we can use the impulse-momentum theorem: the change in momentum equals the impulse delivered by the net force acting during that time. From the lowest point (launch) to the highest point, the only force acting on the projectile is gravity (weight \( mg \), always vertical), and the time taken to reach the highest point is \( t = \dfrac{u\sin\theta}{g} \) (the time for the vertical velocity to drop to zero). The impulse is: \[ J = mg \times t = mg \times \frac{u\sin\theta}{g} = mu\sin\theta. \] Since impulse equals change in momentum, and gravity acts only vertically (horizontal momentum is never touched by any force), this entire change is the net change in momentum. Let's check each option against this result.

1. \( mu \): This would be the full initial momentum magnitude, but momentum is not reduced to zero at the highest point (the horizontal component \( mu\cos\theta \) survives), so this overstates the actual change.
2. \( 2mu \): This would apply only if the velocity fully reversed direction (like an elastic bounce), which does not happen here — the projectile's horizontal motion continues unchanged.
3. \( mu\sin\theta \): This matches exactly the impulse computed from gravity acting over the rise time, confirming this is the actual change in momentum.
4. \( mu\cos\theta \): This is the horizontal momentum, which stays constant throughout the flight (no horizontal force acts) — it represents what does NOT change, not the change itself.

The impulse-momentum theorem confirms the change in momentum from lowest to highest point is \( mu\sin\theta \).

Therefore, the correct answer is \( mu\sin\theta \).