Question

=
=
A particle of mass m = 1 kg is placed with
velocity V = 2 m/s making an angle of 30°
with the horizontal from level ground. When
the particle lands on the level ground the
magnitude of the change in its momentum
will be
1 kg m/s
O
02
C
03 /
2 kg m/s
3 kg m/s
O
4 kg m/s

Answer 1

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Explanation:

We know that in the horizontal plane, the initial velocity of projectile = final velocity of the projectile

=2×1×50×12=50 kg ms−1

Alternative: In the horizontal direction momentum does not change,

Δ Px=0

In vertical direction

ΔPy=(musinθ)−(−musinθ) (Take downward as positive)

=2musinθ downward

=2×1×50×12=50 kg m