Question

A particle of mass m is projected with velocity v

making an angle of 45° with the horizontal. When the

particle lands on the level ground the magnitude of the

change in its momentum will be

(A) 2^1/2(B) zero

(C) 2 (D) / 2^1/2​

Answer 1

Answer:

Hope it helpful dear

Explanation:

Change in momentum =Δ

P

=ΔP

x

i

^

+ΔP

x

j

^

, since ΔP

x

=0

=ΔP

y

j

^

=

P

fy

−

P

iy

;

P

iy

=mvsin45

,P

fy

=mvsin45

Δ

P

=−mvsin45

−mvsin45

=−2mvsin45

=−

2

mv

j

=

∣

∣

∣

∣

Δ

P

∣

∣

∣

∣

=

2

mv

e sign shows the direction of change in

momentum in e y-direction.