The projectile motion of a particle of mass 5 g is shown in the figure.
The initial velocity of the particle is 5√2 ms⁻¹ and the air resistance is assumed to be negligible. The magnitude of the change in momentum between the points A and B is x × 10⁻² kgms⁻¹. The value of x, to the nearest integer, is __________.

Solution

Physics sir · 3 steps

Only the vertical velocity changes from A to B, and it just flips sign, so look there alone.

1. Split the velocity. At A the speed is 5√2 m s⁻¹ at 45°, so uₓ = u_y = 5√2 × 1/√2 = 5 m s⁻¹.
2. Velocity at B. Gravity acts only vertically, so the horizontal component stays 5 m s⁻¹. B is at the same level as A, so the vertical component has the same size but points down: v_y = -5 m s⁻¹.

   |Δ p⃗| = m |v_y - u_y| = 0.005 × |-5 - 5| = 0.05 kg m s⁻¹

3. So. |Δ p⃗| = 5 × 10⁻² kg m s⁻¹, so x = 5.

The common slip: 7. Reversing the whole velocity instead of only its vertical part gives 2 × 0.005 × 5√2 = 7.07 × 10⁻², so 7 instead of 5.

The idea to keep. Between two points at the same level, the change in momentum is 2mu sin θ, straight down. Only the vertical component changes, because gravity is the only force.

Answer from the NTA final answer key. Solution by PrepWiser.