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Step-by-Step Solution
Step 1: Understand the scenario
A particle of mass $m$ is projected with an initial speed $v$ at an angle of 45° above the horizontal. We need to find the change in its momentum from the launch point to the landing point on level ground.
Step 2: Break the initial velocity into components
Since the particle is projected at an angle of 45°, its initial velocity has both horizontal and vertical components:
Horizontal component of velocity: $v_x = v \cos 45^\circ = \frac{v}{\sqrt{2}}$
Vertical component of velocity: $v_y = v \sin 45^\circ = \frac{v}{\sqrt{2}}$
Step 3: Analyze momentum at launch and at landing
Momentum $\vec{p}$ is given by $\vec{p} = m \vec{v}$. Consider the initial and final momenta:
• At the point of projection, the momentum is
$\vec{p}_\text{initial} = m \left( \frac{v}{\sqrt{2}}, \,\frac{v}{\sqrt{2}} \right).$
• When the particle lands on the same level, its speed is the same in magnitude (neglecting air resistance), but the vertical component of velocity is now directed downward:
$\vec{p}_\text{final} = m \left( \frac{v}{\sqrt{2}}, \,-\frac{v}{\sqrt{2}} \right).$
The horizontal component remains unchanged, whereas the vertical component reverses direction.
Step 4: Calculate the change in momentum
The change in momentum $\Delta \vec{p}$ is:
$\Delta \vec{p} = \vec{p}_\text{final} - \vec{p}_\text{initial}.$
Horizontal change: $m\frac{v}{\sqrt{2}} - m\frac{v}{\sqrt{2}} = 0$
Vertical change: $-\,m\frac{v}{\sqrt{2}} - \left(m\frac{v}{\sqrt{2}}\right) = -\,m\frac{v}{\sqrt{2}} - m\frac{v}{\sqrt{2}} = -\,2m\frac{v}{\sqrt{2}}$
Therefore, in vector form,
$\Delta \vec{p} = \left( 0,\,-\,2m\frac{v}{\sqrt{2}} \right).$
Step 5: Find the magnitude of the change in momentum
The magnitude of $\Delta \vec{p}$ is:
$|\Delta \vec{p}| = \sqrt{\left(0\right)^2 + \left(-\,2m\frac{v}{\sqrt{2}}\right)^2} = 2m\frac{v}{\sqrt{2}} = m v\sqrt{2}.$
Thus, the magnitude of the change in momentum is $m v \sqrt{2}$.
Answer
$m v \sqrt{2}$
