Question

A particle of mass m is projected with velocity v making of an angle of 45 degree with the horizontal. when the particle lands on the level ground, then what is the magnitude of the change in its momentum?

Answer 1

Initial momentum=m(vcos45i+vsin45j)
Final momentum=m(vcos45i-vsin45j)
change in momentum=-2mvsin45j
Magnitude of it is 2mvsin45 or √2mv.

Answer 2

The change in momentum is $\bold{\sqrt{2}(m v)}$.

Given:

Mass = m

Velocity = v

Angle = 45°

Solution:

The formula for momentum is given below:

p = mv

From the question, momentum change when the particle reaches ground level is asked.

Thus, the horizontal momentum remains unchanged.

$\Delta p=\Delta p_{v e r t i c a l}$

$\Delta p_{v e r t i c a l}=(\text {Final momentum})_{v e r t i c a l}-(\text {Initial momentum})_{v e r t i c a l}$

Initial momentum = mv sin45

Final momentum = -(mv sin45)

$\Delta p_{v e r t i c a l}=-(m v \sin 45)-(m v \sin 45)$

$=-(2 m v \sin 45)$

$\Delta p_{v e r t i c a l}=2 m v \times \frac{1}{\sqrt{2}}$

$\Delta p_{v e r t i c a l}=\sqrt{2}(m v)$.