Question

a projectile of mass m is fired with velocity v at an angle of 45 with the horizontal from point a. neglecting air resistance,the magnitude of change in momentum from starting point a and the striking point b is?

Answer 1

m = mass, v= initial velocity, projection angle is 45.
Let v' be the striking velocity.
$$\begin{gathered} initialmomentum{P}_i=mv \\ velocityatmaximumheightbecomezero.Accelerationduetogravityg. \\ velocityatstrikingpoint,v{\text{'}}^2-0^2=2gh\left(h=maximumheight\right) \\ h=\frac{v^2{\sin }^2\left({45}^{°}\right)}{2g}=\frac{v^2}{4g} \\ v{\text{'}}^2=\sqrt{2g\left(\frac{v^2}{4g}\right)}=\frac{v}{\sqrt{2}} \\ changeinmomentum=mv-m\frac{v}{\sqrt{2}}=mv\left(\frac{\sqrt{2}-1}{\sqrt{2}}\right) \\ changeinmomentum=0.29mv\left(approx.0.3mv\right) \end{gathered}$$