Question

A particle having a mass $0.5$ kg is projected under gravity with a speed of $98$ m/sec at an angle of $60$ the magnitude of the change of momentum of the particle after $10$ seconds angle of $60$ the magnitude of the change momentum of the particle after $10$ sec is.

Answer 1

m = mass of the particle = 0.5 kg

v₀ = initial velocity of the particle = (98 Cos60) i + (98 Sin60) j = 49 i + 84.87 j

a = acceleration = 0 i - 9.8 j

v = final velocity = ?

t = time of travel = 10 sec

final velocity using the kinematics equation is given as

v = v₀ + at

v = (49 i + 84.87 j ) + (0 i - 9.8 j ) (10)

v = 49 i -13.13 j

change in momentum is given as

ΔP = m (v - v₀)

ΔP = (0.5) ((49 i -13.13 j) - (49 i + 84.87 j))

ΔP = (0.5) (- 98) j

ΔP = - 49 j

magnitude of change in momentum : = 49