A body of mass 0.5 kg is projected under gravity with a speed of 98 m/s at an angle of 30° with the horizontal. The change in momentum (in magnitude) of the body is

The correct answer is 49.0 N–s.

To find the change in magnitude of momentum of the projectile from the point of projection to the point where it returns to the same horizontal level, we analyze the components of its velocity.

Let the initial velocity of projection be $u$ at an angle $\theta$ with the horizontal. The mass of the body is $m$.

The initial velocity vector can be resolved into horizontal and vertical components:
Horizontal component: $u_x=ucos\theta$
Vertical component: $u_y=usin\theta$

When the projectile returns to the same horizontal level, its velocity components are:
Horizontal component: $v_x=ucos\theta$
Vertical component: $v_y=-usin\theta$ (since it is moving downwards)

Now, let us calculate the change in momentum along the horizontal and vertical directions:

Change in horizontal momentum:
$\Delta p_x=m(v_x-u_x)=m(ucos\theta -ucos\theta )=0$

Change in vertical momentum:
$\Delta p_y=m(v_y-u_y)=m(-usin\theta -usin\theta )=-2musin\theta$

Therefore, the magnitude of the net change in momentum is:
$\left|\Delta p\right|=\sqrt{{\left(\Delta p_x\right)}^2+{\left(\Delta p_y\right)}^2}=2musin\theta$

Given values:
Mass, $m=0.5\text{kg}$
Initial speed, $u=98\text{m/s}$
Angle of projection, $\theta =30°$

Substituting the given values into the formula:
$\left|\Delta p\right|=2\times 0.5\times 98\times sin(30°)$

Since $sin(30°)=0.5$:
$\left|\Delta p\right|=1\times 98\times 0.5=49.0\text{N–s}$