Question

a particle of mass 'm' is projected with velocity'v' making an angle of 45 with the horizontal when the particles land on the level ground the magnitude of the change in its momentum will be​

Answer 1

Answer: 0, assuming there's no dissipation.

Explanation:

Since I'm assuming the absence of dissipative forces, the only effective force acting on the particle is the gravitational force due to Earth.

Applying the Principle of conservation of Mechanical Energy, we know that

ΔK= -ΔU, i.e, the loss in potential energy is compensated by an equal increase in kinetic energy and vice versa.

In this case, the particle is launched from the surface of the Earth. When the particle hits the surface of the Earth again, its potential energy remains unchanged ⇒ΔU=0 ⇒ΔK=0, i.e., the kinetic energy of the particle remains unchanged as well. By definition,

$K=\frac{1}{2} mv^2=\frac{1}{2} v(mv)=\frac{pv}{2m}$

As shown in the above equation, the K.E of a particle is directly proportional to the magnitude of its momentum. Since ΔK=0, ∴, Δp=0, i.e., the change in magnitude of momentum is 0.

Answer 2

Answer:

$\sqrt{2}$ mv

Explanation:

Here the particle follows a projectile motion.

In projectile Motion the velocity along the x-axis does not change.

∴ Change in momentum along x-axis is zero.

we will only consider the velocity along the y-axis to get the change in momentum of the particle.

so,

velocity along the y-axis = vsin45°

|ΔP|= m(Δv)

|ΔP|= m[vsin45°-(-vsin45°)]

|ΔP|=2×mv×sin45°

|ΔP|=2×mv×1/$\sqrt{2}$

|ΔP|= $\sqrt{2}$mv