Question

A body of mass $m$ is projected at an angle of $${45}^{\circ }$$ with the horizontal. If air resistance is negligible, then what will be the total change in momentum when it strikes the ground?

Answer 1

Hint: In this solution, we will first calculate the component of the projectile body perpendicular to the surface of the Earth when it strikes the ground. The ball after striking the ground will maintain the magnitude of its velocity but reverse its direction.

Complete step by step answer:
We’ve been told a projectile is launched at an angle of $${45}^{\circ }$$ with the horizontal. When the projectile is about to complete its motion and collide back with the ground, all its energy will be in the form of kinetic energy when it strikes the ground. This implies that the velocity of the projectile will have the same value when it was launched.
However, the perpendicular component of the velocity will be downwards since the ball will be approaching the ground. If the projectile of the velocity is denoted by $v$, its component perpendicular to the ground will be $v\sin \theta$.
Now when the ball strikes the ground it will bounce back up. But this time, it will have a perpendicular velocity in the upwards direction that is opposite to its initial direction when it strikes the ground. So, we can calculate the net change in momentum as
$\mathrm{\Delta }P=mv\sin \theta -(-mv\sin \theta )$
$\Rightarrow \mathrm{\Delta }P=2mv\sin \theta$
Since $$\theta ={45}^{\circ }$$,
$\mathrm{\Delta }P=2mv{\displaystyle \frac{1}{\sqrt{2}}}$
Which can be simplified to
$\mathrm{\Delta }P={\displaystyle \frac{mv}{\sqrt{2}}}$

Note: Here we have assumed that the ball will not lose its energy when it collides with the ground. In reality, there will always be a loss of energy when the ball collides with the ground and hence its velocity will be lower than the initial velocity it had when striking the ground.