Solveeit Logo

Question

Question: A ball of mass m is projected from the ground with a speed u at an angle with the horizontal. The ma...

A ball of mass m is projected from the ground with a speed u at an angle with the horizontal. The magnitude of the change in momentum of the ball over a time interval from beginning till it strikes the ground again is
A. musinα2\dfrac{{mu\sin \alpha }}{2}
B. 2mucosα2mu\cos \alpha
C. mucosα2\dfrac{{mu\cos \alpha }}{2}
D. 2musinα2mu\sin \alpha

Explanation

Solution

Resolve the velocity component and hence the momentum in vertical and horizontal components. One of the components remains the same while the other changes. Calculate the changed momentum and the final change in the momentum will be equal to the sum of the change in momentum along the two components.

Complete step by step answer:

The above figure depicts the trajectory of the ball from the start point to the point where it strikes back to the ground. From the figure we note that the x component of the velocity remains same i.e.
uxi=uxf=ucosαu_x^i = u_x^f = u\cos \alpha
So, the momentum will be same along x direction and there will be no change in momentum
Δpx=0\Delta {p_x} = 0

The change in momentum in y direction can be calculated as
pyi=musinα     pyf=musinα     Δpy=pyfpyi=musinαmusinα     Δpy=2musinα  p_y^i = mu\sin \alpha \\\ \implies p_y^f = - mu\sin \alpha \\\ \implies \Delta {p_y} = p_y^f - p_y^i = - mu\sin \alpha - mu\sin \alpha \\\ \implies \Delta {p_y} = - 2mu\sin \alpha \\\
So, the net change in momentum is
Δp=Δpx+Δpy     Δp=02musinα=2musinα     Δp=2musinα  \Delta p = \Delta {p_x} + \Delta {p_y} \\\ \implies \Delta p = 0 - 2mu\sin \alpha = - 2mu\sin \alpha \\\ \implies \left| {\Delta p} \right| = 2mu\sin \alpha \\\
Here we have ignored the negative sign depending upon the options provided in the question. It depends upon the perspective of the person and hence the sign convention may change. Therefore I have taken the absolute value of Δp\Delta p in the last step. The problem maker has taken a positive y direction opposite to what is used here.

So, the correct answer is “Option D”.

Note:
Always remember that in 2D trajectory motion like the one depicted above, one component of velocity always remains the same at the start and end of the path. Mostly it is the x-component in the general coordinate convention method.