Question

An object of mass 7.20 kg is projected into the air at a 55.0° angle. It hits the ground 3.40 s later. Set "up" to be the positive y-direction. What is the y-component of the object's change in momentum while it is in the air? Ignore air resistance.

Answer 1

Given,

The mass of the object, m=7.20 kg

The angle of projection, θ=55.0°

The time of flight, t=3.40 s

The time of flight of a projectile is given by,

`t=(2u* sin\theta)/(g)`

Where u is the initial velocity and g is the acceleration due to gravity.

On substituting the known values,

`\begin{gathered} 3.40=(2* u* sin55^(\circ))/(9.8) \\ \Rightarrow u=(3.40*9.8)/(2* sin55^(\circ)) \\ =20.34\text{ m/s} \end{gathered}`

Thus the y-component of the initial momentum is,

`p_(iy)=mu\sin \theta`

On substituting the known values,

`\begin{gathered} p_(iy)=7.20*20.34*\sin 55^(\circ) \\ =119.96\text{ kg}\cdot(m)/(s) \end{gathered}`

The y-component of the final velocity can be calculated using one of the equations of the motion,

`v_y=u_{}\sin \theta-gt`

On substituting the known values,

`\begin{gathered} v_y=20.34*\sin 55^(\circ)-9.8*3.40 \\ =-16.66\text{ m/s} \end{gathered}`

Thus the y-component of the final momentum is

`p_(fy)=mv_y_{}`

On substituting the known values,

`\begin{gathered} p_(fy)=7.20*-16.66 \\ =-119.95\text{ kg}\cdot(m)/(s) \end{gathered}`

The y-component of the change in the momentum is given by,

`\Delta p_y=p_(fy)-p_(iy)`

On substituting the known values,

`\begin{gathered} \Delta p_y=-119.95-119.96 \\ =-239.91\text{ kg}\cdot(m)/(s) \end{gathered}`

Thus the y-component of the object's change in momentum is -239.91 kg·m/s

The negative sign indicates that the change is directed downwards.