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: No horizontal force acts on the projectile, so the horizontal momentum \( mu\cos\theta \) is conserved and contributes zero to any change.
Step 2: The vertical velocity changes from \( u\sin\theta \) at launch to \( 0 \) at the highest point, so the vertical momentum changes by \( m(u\sin\theta - 0) = mu\sin\theta \).
Step 3: Since only the vertical component changes, the net change in momentum is entirely this vertical change.
\[ \boxed{\Delta p = mu\sin\theta} \]

Answer 2

We can also find the change in momentum by directly subtracting the momentum vectors at the two points. At the lowest point, the velocity vector is \( (u\cos\theta, u\sin\theta) \), so momentum is \( (mu\cos\theta, mu\sin\theta) \). At the highest point, the velocity vector is \( (u\cos\theta, 0) \), so momentum is \( (mu\cos\theta, 0) \). The change in momentum vector is: \[ \Delta \vec{p} = (mu\cos\theta - mu\cos\theta, \, 0 - mu\sin\theta) = (0, -mu\sin\theta), \] whose magnitude is simply \( mu\sin\theta \). Let's check each option against this vector subtraction.

1. \( mu \): This does not appear anywhere in the subtracted vector; the horizontal components cancel entirely and only the vertical term \( mu\sin\theta \) survives, not the full magnitude \( mu \).
2. \( 2mu \): This would require both components to reverse and double, which is not what the vector subtraction shows here.
3. \( mu\sin\theta \): This is exactly the magnitude of the resulting difference vector \( (0, -mu\sin\theta) \).
4. \( mu\cos\theta \): This term cancels out completely in the subtraction (it's identical at both points), so it cannot be part of the change.

Direct vector subtraction of the momentum at the two points confirms the change is \( mu\sin\theta \).

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