A particle of mass 100 g is fired with a velocity 20 m sec⁻¹ making an angle of 30° with the horizontal. When it rises to the highest point of its path then the change in its momentum is

The correct option is 1 kg m sec⁻¹.

Let us solve the problem step-by-step.

1. Identify the given parameters:
Mass of the particle, m = 100 g = 0.1 kg
Initial velocity, u = 20 m sec⁻¹
Angle of projection with the horizontal, θ = 30°

2. Express the initial velocity in vector form:
At the point of projection, the velocity has both horizontal and vertical components:
The horizontal component of velocity is:
$u_x=u\cos \theta$
The vertical component of velocity is:
$u_y=u\sin \theta$
Therefore, the initial velocity vector is:
${\overrightarrow{v}}_i=\left(u\cos \theta \right)\hat{i}+\left(u\sin \theta \right)\hat{j}$
Initial momentum vector is:
${\overrightarrow{p}}_i=m{\overrightarrow{v}}_i=m\left(u\cos \theta \right)\hat{i}+m\left(u\sin \theta \right)\hat{j}$

3. Determine the velocity and momentum at the highest point:
At the highest point of projectile motion, the vertical component of velocity becomes zero ($v_y=0$). Only the horizontal component remains, which is constant throughout the motion because there is no horizontal acceleration.
Thus, the velocity vector at the highest point is:
${\overrightarrow{v}}_f=\left(u\cos \theta \right)\hat{i}$
The momentum vector at the highest point is:
${\overrightarrow{p}}_f=m{\overrightarrow{v}}_f=m\left(u\cos \theta \right)\hat{i}$

4. Calculate the change in momentum:
The change in momentum is given by:
$\Delta \overrightarrow{p}={\overrightarrow{p}}_f-{\overrightarrow{p}}_i$
Substitute the values of final and initial momentum:
$\Delta \overrightarrow{p}=m\left(u\cos \theta \right)\hat{i}-\left[m\left(u\cos \theta \right)\hat{i}+m\left(u\sin \theta \right)\hat{j}\right]$
$\Delta \overrightarrow{p}=-m\left(u\sin \theta \right)\hat{j}$
Taking the magnitude of this change:
$\left|\Delta \overrightarrow{p}\right|=mu\sin \theta$

5. Substitute the numerical values:
Substitute m = 0.1 kg, u = 20 m sec⁻¹, and θ = 30°:
$\left|\Delta \overrightarrow{p}\right|=0.1\times 20\times \sin (30°)$
Since $\sin (30°)=\frac{1}{2}$:
$\left|\Delta \overrightarrow{p}\right|=2\times \frac{1}{2}=1\text{kg m sec}{⁻}^1$

Thus, the change in momentum of the particle when it reaches its highest point is 1 kg m sec⁻¹.